A FIRST COURSE IN PROBABILITY

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A FIRST COURSE IN PROBABILITY

Eighth Edition

Sheldon Ross

University of Southern California

Upper Saddle River, New Jersey 07458

Library of Congress Cataloging-in-Publication Data

Ross, Sheldon M.

A ﬁrst course in probability / Sheldon Ross. — 8th ed.

p. cm.

Includes bibliographical references and index.

ISBN-13: 978-0-13-603313-4

ISBN-10: 0-13-603313-X

1. Probabilities—Textbooks. I. Title.

QA273.R83 2010

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2008033720

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Contents

Preface xi

1 Combinatorial Analysis 1

1.1 Introduction . . . .............................. 1

1.2 The Basic Principle of Counting . . . ................... 1

1.3 Permutations................................. 3

1.4 Combinations . . .............................. 5

1.5 Multinomial Coefﬁcients . . . ....................... 9

1.6 The Number of Integer Solutions of Equations . ............ 12

Summary . .................................. 15

Problems................................... 16

Theoretical Exercises . . . . . ....................... 18

Self-Test Problems and Exercises . . ................... 20

2 Axioms of Probability 22

2.1 Introduction . . . .............................. 22

2.2 SampleSpaceandEvents.......................... 22

2.3 Axioms of Probability . . . . . ....................... 26

2.4 Some Simple Propositions . . ....................... 29

2.5 Sample Spaces Having Equally Likely Outcomes ............ 33

2.6 Probability as a Continuous Set Function . . . . . ............ 44

2.7 Probability as a Measure of Belief . . ................... 48

Summary . .................................. 49

Problems................................... 50

Theoretical Exercises . . . . . ....................... 54

Self-Test Problems and Exercises . . ................... 56

3 Conditional Probability and Independence 58

3.1 Introduction . . . .............................. 58

3.2 Conditional Probabilities . . . ....................... 58

3.3 Bayes’s Formula . .............................. 65

3.4 IndependentEvents............................. 79

3.5 P(·|F) Is a Probability . . . . . ....................... 93

Summary . .................................. 101

Problems................................... 102

Theoretical Exercises . . . . . ....................... 110

Self-Test Problems and Exercises . . ................... 114

4 Random Variables 117

4.1 Random Variables .............................. 117

4.2 Discrete Random Variables . ....................... 123

4.3 ExpectedValue ............................... 125

4.4 Expectation of a Function of a Random Variable ............ 128

4.5 Variance . .................................. 132

4.6 The Bernoulli and Binomial Random Variables . ............ 134

4.6.1 Properties of Binomial Random Variables ............ 139

4.6.2 Computing the Binomial Distribution Function . . . . ..... 142

vii

viii Contents

4.7 The Poisson Random Variable ....................... 143

4.7.1 Computing the Poisson Distribution Function . . . . . ..... 154

4.8 Other Discrete Probability Distributions . . . . . ............ 155

4.8.1 The Geometric Random Variable . . . . . ............ 155

4.8.2 The Negative Binomial Random Variable ............ 157

4.8.3 The Hypergeometric Random Variable . ............ 160

4.8.4 TheZeta(orZipf)Distribution.................. 163

4.9 Expected Value of Sums of Random Variables . ............ 164

4.10 Properties of the Cumulative Distribution Function . . . . . . ..... 168

Summary . .................................. 170

Problems................................... 172

Theoretical Exercises . . . . . ....................... 179

Self-Test Problems and Exercises . . ................... 183

5 Continuous Random Variables 186

5.1 Introduction . . . .............................. 186

5.2 Expectation and Variance of Continuous Random Variables ..... 190

5.3 The Uniform Random Variable . . . ................... 194

5.4 Normal Random Variables . . ....................... 198

5.4.1 The Normal Approximation to the Binomial Distribution . . . 204

5.5 Exponential Random Variables . . . ................... 208

5.5.1 Hazard Rate Functions ....................... 212

5.6 Other Continuous Distributions . . . ................... 215

5.6.1 TheGammaDistribution ..................... 215

5.6.2 TheWeibullDistribution ..................... 216

5.6.3 TheCauchyDistribution...................... 217

5.6.4 TheBetaDistribution ....................... 218

5.7 The Distribution of a Function of a Random Variable . . . . ..... 219

Summary . .................................. 222

Problems................................... 224

Theoretical Exercises . . . . . ....................... 227

Self-Test Problems and Exercises . . ................... 229

6 Jointly Distributed Random Variables 232

6.1 JointDistributionFunctions ........................ 232

6.2 Independent Random Variables . . . ................... 240

6.3 Sums of Independent Random Variables . . . . . ............ 252

6.3.1 Identically Distributed Uniform Random Variables . ..... 252

6.3.2 Gamma Random Variables . ................... 254

6.3.3 Normal Random Variables . ................... 256

6.3.4 Poisson and Binomial Random Variables ............ 259

6.3.5 Geometric Random Variables ................... 260

6.4 Conditional Distributions: Discrete Case . . . . . ............ 263

6.5 Conditional Distributions: Continuous Case . . . ............ 266

6.6 OrderStatistics ............................... 270

6.7 Joint Probability Distribution of Functions of Random Variables . . . 274

6.8 Exchangeable Random Variables . . ................... 282

Summary . .................................. 285

Problems................................... 287

Theoretical Exercises . . . . . ....................... 291

Self-Test Problems and Exercises . . ................... 293

Contents ix

7 Properties of Expectation 297

7.1 Introduction . . . .............................. 297

7.2 Expectation of Sums of Random Variables . . . . ............ 298

7.2.1 Obtaining Bounds from Expectations

via the Probabilistic Method . ................... 311

7.2.2 The Maximum–Minimums Identity . . . . ............ 313

7.3 Moments of the Number of Events that Occur . . ............ 315

7.4 Covariance, Variance of Sums, and Correlations . ............ 322

7.5 Conditional Expectation . . . ....................... 331

7.5.1 Deﬁnitions.............................. 331

7.5.2 Computing Expectations by Conditioning ............ 333

7.5.3 Computing Probabilities by Conditioning ............ 344

7.5.4 Conditional Variance . ....................... 347

7.6 Conditional Expectation and Prediction . . . . . ............ 349

7.7 Moment Generating Functions ....................... 354

7.7.1 Joint Moment Generating Functions . . . ............ 363

7.8 Additional Properties of Normal Random Variables . . . . . ..... 365

7.8.1 The Multivariate Normal Distribution . . ............ 365

7.8.2 The Joint Distribution of the Sample Mean

and Sample Variance . ....................... 367

7.9 General Deﬁnition of Expectation . . ................... 369

Summary . .................................. 370

Problems................................... 373

Theoretical Exercises . . . . . ....................... 380

Self-Test Problems and Exercises . . ................... 384

8 Limit Theorems 388

8.1 Introduction . . . .............................. 388

8.2 Chebyshev’s Inequality and the Weak Law of Large

Numbers . .................................. 388

8.3 TheCentralLimitTheorem ........................ 391

8.4 The Strong Law of Large Numbers . ................... 400

8.5 Other Inequalities .............................. 403

8.6 Bounding the Error Probability When Approximating a Sum of

Independent Bernoulli Random Variables by a Poisson

Random Variable .............................. 410

Summary . .................................. 412

Problems................................... 412

Theoretical Exercises . . . . . ....................... 414

Self-Test Problems and Exercises . . ................... 415

9 Additional Topics in Probability 417

9.1 The Poisson Process . . . . . . ....................... 417

9.2 MarkovChains................................ 419

9.3 Surprise, Uncertainty, and Entropy . ................... 425

9.4 Coding Theory and Entropy . ....................... 428

Summary . .................................. 434

Problems and Theoretical Exercises . ................... 435

Self-Test Problems and Exercises . . ................... 436

References .................................. 436

x Contents

10 Simulation 438

10.1 Introduction . . . .............................. 438

10.2 General Techniques for Simulating Continuous Random Variables . . 440

10.2.1 The Inverse Transformation Method . . . ............ 441

10.2.2 The Rejection Method ....................... 442

10.3 Simulating from Discrete Distributions . . . . . . ............ 447

10.4 Variance Reduction Techniques . . . ................... 449

10.4.1 Use of Antithetic Variables . ................... 450

10.4.2 Variance Reduction by Conditioning . . . ............ 451

10.4.3 Control Variates . . . ....................... 452

Summary . .................................. 453

Problems................................... 453

Self-Test Problems and Exercises . . ................... 455

Reference .................................. 455

Answers to Selected Problems 457

Solutions to Self-Test Problems and Exercises 461

Index 521

Preface

“We see that the theory of probability is at bottom only common sense reduced

to calculation; it makes us appreciate with exactitude what reasonable minds feel

by a sort of instinct, often without being able to account for it. ...It is remarkable

that this science, which originated in the consideration of games of chance, should

have become the most important object of human knowledge. ...The most important

questions of life are, for the most part, really only problems of probability.” So said

the famous French mathematician and astronomer (the “Newton of France”) Pierre-

Simon, Marquis de Laplace. Although many people feel that the famous marquis,

who was also one of the great contributors to the development of probability, might

have exaggerated somewhat, it is nevertheless true that probability theory has become

a tool of fundamental importance to nearly all scientists, engineers, medical practi-

tioners, jurists, and industrialists. In fact, the enlightened individual had learned to

ask not “Is it so?” but rather “What is the probability that it is so?”

This book is intended as an elementary introduction to the theory of probability

for students in mathematics, statistics, engineering, and the sciences (including com-

puter science, biology, the social sciences, and management science) who possess the

prerequisite knowledge of elementary calculus. It attempts to present not only the

mathematics of probability theory, but also, through numerous examples, the many

diverse possible applications of this subject.

Chapter 1 presents the basic principles of combinatorial analysis, which are most

useful in computing probabilities.

Chapter 2 handles the axioms of probability theory and shows how they can be

applied to compute various probabilities of interest.

Chapter 3 deals with the extremely important subjects of conditional probability

and independence of events. By a series of examples, we illustrate how conditional

probabilities come into play not only when some partial information is available,

but also as a tool to enable us to compute probabilities more easily, even when

no partial information is present. This extremely important technique of obtaining

probabilities by “conditioning” reappears in Chapter 7, where we use it to obtain

expectations.

The concept of random variables is introduced in Chapters 4, 5, and 6. Discrete

random variables are dealt with in Chapter 4, continuous random variables in

Chapter 5, and jointly distributed random variables in Chapter 6. The important con-

cepts of the expected value and the variance of a random variable are introduced in

Chapters 4 and 5, and these quantities are then determined for many of the common

types of random variables.

Additional properties of the expected value are considered in Chapter 7. Many

examples illustrating the usefulness of the result that the expected value of a sum of

random variables is equal to the sum of their expected values are presented. Sections

on conditional expectation, including its use in prediction, and on moment-generating

functions are contained in this chapter. In addition, the ﬁnal section introduces the

multivariate normal distribution and presents a simple proof concerning the joint

distribution of the sample mean and sample variance of a sample from a normal

distribution.

xii Preface

Chapter 8 presents the major theoretical results of probability theory. In partic-

ular, we prove the strong law of large numbers and the central limit theorem. Our

proof of the strong law is a relatively simple one which assumes that the random vari-

ables have a ﬁnite fourth moment, and our proof of the central limit theorem assumes

Levy’s continuity theorem. This chapter also presents such probability inequalities as

Markov’s inequality, Chebyshev’s inequality, and Chernoff bounds. The ﬁnal section

of Chapter 8 gives a bound on the error involved when a probability concerning

a sum of independent Bernoulli random variables is approximated by the corre-

sponding probability of a Poisson random variable having the same expected

value.

Chapter 9 presents some additional topics, such as Markov chains, the Poisson pro-

cess, and an introduction to information and coding theory, and Chapter 10 considers

simulation.

As in the previous edition, three sets of exercises are given at the end of each

chapter. They are designated as Problems, Theoretical Exercises,andSelf-Test Prob-

lems and Exercises. This last set of exercises, for which complete solutions appear in

Solutions to Self-Test Problems and Exercises, is designed to help students test their

comprehension and study for exams.

CHANGES IN THE NEW EDITION

The eighth edition continues the evolution and ﬁne tuning of the text. It includes

new problems, exercises, and text material chosen both for its inherent interest and

for its use in building student intuition about probability. Illustrative of these goals

are Example 5d of Chapter 1 on knockout tournaments, and Examples 4k and 5i of

Chapter 7 on multiple player gambler’s ruin problems.

A key change in the current edition is that the important result that the expectation

of a sum of random variables is equal to the sum of the expectations is now ﬁrst

presented in Chapter 4 (rather than Chapter 7 as in previous editions). A new and

elementary proof of this result when the sample space of the probability experiment

is ﬁnite is given in this chapter.

Another change is the expansion of Section 6.3, which deals with the sum of inde-

pendent random variables. Section 6.3.1 is a new section in which we derive the

distribution of the sum of independent and identically distributed uniform random

variables, and then use our results to show that the expected number of random num-

bers that needs to be added for their sum to exceed 1 is equal to e. Section 6.3.5 is a

new section in which we derive the distribution of the sum of independent geometric

random variables with different means.

ACKNOWLEDGMENTS

I am grateful for the careful work of Hossein Hamedani in checking the text for accu-

racy. I also appreciate the thoughtfulness of the following people that have taken the

time to contact me with comments for improving the text: Amir Ardestani, Polytech-

nic University of Teheran; Joe Blitzstein, Harvard University; Peter Nuesch, Univer-

sity of Lausaunne; Joseph Mitchell, SUNY, Stony Brook; Alan Chambless, actuary;

Robert Kriner; Israel David, Ben-Gurion University; T. Lim, George Mason Univer-

sity; Wei Chen, Rutgers; D. Monrad, University of Illinois; W. Rosenberger, George

Mason University; E. Ionides, University of Michigan; J. Corvino, Lafayette College;

T. Seppalainen, University of Wisconsin.

Finally, I would like to thank the following reviewers for their many helpful com-

ments. Reviewers of the eighth edition are marked with an asterisk.

Acknowledgments xiii

K. B. Athreya, Iowa State University

Richard Bass, University of Connecticut

Robert Bauer, University of Illinois at

Urbana-Champaign

Phillip Beckwith, Michigan Tech

Arthur Benjamin, Harvey Mudd College

Geoffrey Berresford, Long Island University

Baidurya Bhattacharya, University of Delaware

Howard Bird, St. Cloud State University

Shahar Boneh, Metropolitan State College of

Denver

Jean Cadet, State University of New York at

Stony Brook

Steven Chiappari, Santa Clara University

Nicolas Christou, University of California, Los

Angeles

James Clay, University of Arizona at Tucson

Francis Conlan, University of Santa Clara

*Justin Corvino, Lafayette College

Jay DeVore, California Polytechnic University,

San Luis Obispo

Scott Emerson, University of Washington

Thomas R. Fischer, Texas A & M University

Anant Godbole, Michigan Technical

University

Zakkula Govindarajulu, University of Kentucky

Richard Groeneveld, Iowa State University

Mike Hardy, Massachusetts Institute of

Technology

Bernard Harris, University of Wisconsin

Larry Harris, University of Kentucky

David Heath, Cornell University

Stephen Herschkorn, Rutgers University

Julia L. Higle, University of Arizona

Mark Huber, Duke University

*Edward Ionides, University of Michigan

Anastasia Ivanova, University of North Carolina

Hamid Jafarkhani, University of California,

Irvine

Chuanshu Ji, University of North Carolina,

Chapel Hill

Robert Keener, University of Michigan

Fred Leysieffer, Florida State University

Thomas Liggett, University of California, Los

Angeles

Helmut Mayer, University of Georgia

Bill McCormick, University of Georgia

Ian McKeague, Florida State University

R. Miller, Stanford University

*Ditlev Monrad, University of Illinois

Robb J. Muirhead, University of Michigan

Joe Naus, Rutgers University

Nhu Nguyen, New Mexico State University

Ellen O’Brien, George Mason University

N. U. Prabhu, Cornell University

Kathryn Prewitt,

Arizona State University

Jim Propp, University of Wisconsin

*William F. Rosenberger, George Mason

University

Myra Samuels, Purdue University

I. R. Savage, Yale University

Art Schwartz, University of Michigan at Ann

Arbor

Therese Shelton, Southwestern University

Malcolm Sherman, State University of New York

at Albany

Murad Taqqu, Boston University

Eli Upfal, Brown University

Ed Wheeler, University of Tennessee

Allen Webster, Bradley University

S. R.

smross@usc.edu

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CHAPTER 1

Combinatorial Analysis

1.1 INTRODUCTION

1.2 THE BASIC PRINCIPLE OF COUNTING

1.3 PERMUTATIONS

1.4 COMBINATIONS

1.5 MULTINOMIAL COEFFICIENTS

1.6 THE NUMBER OF INTEGER SOLUTIONS OF EQUATIONS

1.1 INTRODUCTION

Here is a typical problem of interest involving probability: A communication system

is to consist of n seemingly identical antennas that are to be lined up in a linear order.

The resulting system will then be able to receive all incoming signals—and will be

called functional—as long as no two consecutive antennas are defective. If it turns

out that exactly m of the n antennas are defective, what is the probability that the

resulting system will be functional? For instance, in the special case where n = 4and

m = 2, there are 6 possible system conﬁgurations, namely,

0110

0101

1010

0011

1001

1100

where 1 means that the antenna is working and 0 that it is defective. Because the

resulting system will be functional in the ﬁrst 3 arrangements and not functional in

the remaining 3, it seems reasonable to take

as the desired probability. In

the case of general n and m, we could compute the probability that the system is

functional in a similar fashion. That is, we could count the number of conﬁgurations

that result in the system’s being functional and then divide by the total number of all

possible conﬁgurations.

From the preceding discussion, we see that it would be useful to have an effective

method for counting the number of ways that things can occur. In fact, many prob-

lems in probability theory can be solved simply by counting the number of different

ways that a certain event can occur. The mathematical theory of counting is formally

known as combinatorial analysis.

1.2 THE BASIC PRINCIPLE OF COUNTING

The basic principle of counting will be fundamental to all our work. Loosely put, it

states that if one experiment can result in any of m possible outcomes and if another

experiment can result in any of n possible outcomes, then there are mn possible out-

comes of the two experiments.

2 Chapter 1 Combinatorial Analysis

The basic principle of counting

Suppose that two experiments are to be performed. Then if experiment 1 can result

in any one of m possible outcomes and if, for each outcome of experiment 1, there

are n possible outcomes of experiment 2, then together there are mn possible out-

comes of the two experiments.

Proof of the Basic Principle: The basic principle may be proven by enumerating all

the possible outcomes of the two experiments; that is,

(1, 1), (1, 2), ..., (1, n)

(2, 1), (2, 2), ..., (2, n)

(m,1), (m,2), ..., (m, n)

where we say that the outcome is (i, j ) if experiment 1 results in its ith possible out-

come and experiment 2 then results in its jth possible outcome. Hence, the set of

possible outcomes consists of m rows, each containing n elements. This proves the

result.

EXAMPLE 2a

A small community consists of 10 women, each of whom has 3 children. If one woman

and one of her children are to be chosen as mother and child of the year, how many

different choices are possible?

Solution. By regarding the choice of the woman as the outcome of the ﬁrst experi-

ment and the subsequent choice of one of her children as the outcome of the second

experiment, we see from the basic principle that there are 10 * 3 = 30 possible

choices. .

When there are more than two experiments to be performed, the basic principle

can be generalized.

The generalized basic principle of counting

If r experiments that are to be performed are such that the ﬁrst one may result in

any of n

possible outcomes; and if, for each of these n

possible outcomes, there

are n

possible outcomes of the second experiment; and if, for each of the possible

outcomes of the ﬁrst two experiments, there are n

possible outcomes of the third

experiment; and if ..., then there is a total of n

· n

···n

possible outcomes of the

r experiments.

EXAMPLE 2b

A college planning committee consists of 3 freshmen, 4 sophomores, 5 juniors, and 2

seniors. A subcommittee of 4, consisting of 1 person from each class, is to be chosen.

How many different subcommittees are possible?

Section 1.3 Permutations 3

Solution. We may regard the choice of a subcommittee as the combined outcome of

the four separate experiments of choosing a single representative from each of the

classes. It then follows from the generalized version of the basic principle that there

are 3 * 4 * 5 * 2 = 120 possible subcommittees. .

EXAMPLE 2c

How many different 7-place license plates are possible if the ﬁrst 3 places are to be

occupied by letters and the ﬁnal 4 by numbers?

Solution. By the generalized version of the basic principle, the answer is 26 · 26 ·

26 · 10 · 10 · 10 · 10 = 175,760,000. .

EXAMPLE 2d

How many functions deﬁned on n points are possible if each functional value is either

0or1?

Solution. Let the points be 1, 2, ..., n. Since f (i) must be either 0 or 1 for each i =

1, 2, ..., n, it follows that there are 2

possible functions. .

EXAMPLE 2e

In Example 2c, how many license plates would be possible if repetition among letters

or numbers were prohibited?

Solution. In this case, there would be 26 · 25 · 24 · 10 · 9 · 8 · 7 = 78,624,000

possible license plates. .

1.3 PERMUTATIONS

How many different ordered arrangements of the letters a, b,andc are possible? By

direct enumeration we see that there are 6, namely, abc, acb, bac, bca, cab,andcba.

Each arrangement is known as a permutation. Thus, there are 6 possible permutations

of a set of 3 objects. This result could also have been obtained from the basic principle,

since the ﬁrst object in the permutation can be any of the 3, the second object in the

permutation can then be chosen from any of the remaining 2, and the third object

in the permutation is then the remaining 1. Thus, there are 3 · 2 · 1 = 6 possible

permutations.

Suppose now that we have n objects. Reasoning similar to that we have just used

for the 3 letters then shows that there are

n(n − 1)(n − 2) ···3 · 2 · 1 = n!

different permutations of the n objects.

EXAMPLE 3a

How many different batting orders are possible for a baseball team consisting of 9

players?

Solution. There are 9! = 362,880 possible batting orders. .

4 Chapter 1 Combinatorial Analysis

EXAMPLE 3b

A class in probability theory consists of 6 men and 4 women. An examination is given,

and the students are ranked according to their performance. Assume that no two

students obtain the same score.

(a) How many different rankings are possible?

(b) If the men are ranked just among themselves and the women just among them-

selves, how many different rankings are possible?

Solution. (a) Because each ranking corresponds to a particular ordered arrangement

of the 10 people, the answer to this part is 10! = 3,628,800.

(b) Since there are 6! possible rankings of the men among themselves and 4! possi-

ble rankings of the women among themselves, it follows from the basic principle that

there are (6!)(4!) = (720)(24) = 17,280 possible rankings in this case. .

EXAMPLE 3c

Ms. Jones has 10 books that she is going to put on her bookshelf. Of these, 4 are

mathematics books, 3 are chemistry books, 2 are history books, and 1 is a language

book. Ms. Jones wants to arrange her books so that all the books dealing with the

same subject are together on the shelf. How many different arrangements are

possible?

Solution. There are 4! 3! 2! 1! arrangements such that the mathematics books are

ﬁrst in line, then the chemistry books, then the history books, and then the language

book. Similarly, for each possible ordering of the subjects, there are 4! 3! 2! 1! possible

arrangements. Hence, as there are 4! possible orderings of the subjects, the desired

answer is 4! 4! 3! 2! 1! = 6912. .

We shall now determine the number of permutations of a set of n objects when cer-

tain of the objects are indistinguishable from each other. To set this situation straight

in our minds, consider the following example.

EXAMPLE 3d

How many different letter arrangements can be formed from the letters PEPPER?

Solution. We ﬁrst note that there are 6! permutations of the letters P

when the 3P’s and the 2E’s are distinguished from each other. However, consider

any one of these permutations—for instance, P

R. If we now permute the

P’s among themselves and the E’s among themselves, then the resultant arrangement

would still be of the form PPEPER. That is, all 3! 2! permutations

are of the form PPEPER. Hence, there are 6!/(3! 2!) = 60 possible letter arrange-

ments of the letters PEPPER. .

Section 1.4 Combinations 5

In general, the same reasoning as that used in Example 3d shows that there are

! n

! ··· n

different permutations of n objects, of which n

are alike, n

are alike, ..., n

are

alike.

EXAMPLE 3e

A chess tournament has 10 competitors, of which 4 are Russian, 3 are from the United

States, 2 are from Great Britain, and 1 is from Brazil. If the tournament result lists just

the nationalities of the players in the order in which they placed, how many outcomes

are possible?

Solution. There are

10!

4! 3! 2! 1!

= 12,600

possible outcomes. .

EXAMPLE 3f

How many different signals, each consisting of 9 ﬂags hung in a line, can be made

from a set of 4 white ﬂags, 3 red ﬂags, and 2 blue ﬂags if all ﬂags of the same color are

identical?

Solution. There are

4! 3! 2!

= 1260

different signals. .

1.4 COMBINATIONS

We are often interested in determining the number of different groups of r objects

that could be formed from a total of n objects. For instance, how many different

groups of 3 could be selected from the 5 items A, B, C, D,andE? To answer this

question, reason as follows: Since there are 5 ways to select the initial item, 4 ways to

then select the next item, and 3 ways to select the ﬁnal item, there are thus 5 · 4 · 3

ways of selecting the group of 3 when the order in which the items are selected is

relevant. However, since every group of 3—say, the group consisting of items A, B,

and C—will be counted 6 times (that is, all of the permutations ABC, ACB, BAC,

BCA, CAB,andCBA will be counted when the order of selection is relevant), it

follows that the total number of groups that can be formed is

5 · 4 · 3

3 · 2 · 1

= 10

In general, as n(n − 1) ···(n − r + 1) represents the number of different ways that a

group of r items could be selected from n items when the order of selection is relevant,

and as each group of r items will be counted r ! times in this count, it follows that the

number of different groups of r items that could be formed from a set of n items is

n(n − 1) ···(n − r + 1)

(n − r)! r!

6 Chapter 1 Combinatorial Analysis

Notation and terminology

We deﬁne





,forr … n,by





(n − r)! r!

and say that





represents the number of possible combinations of n objects

taken r at a time.

†

Thus,





represents the number of different groups of size r that could be

selected from a set of n objects when the order of selection is not considered relevant.

EXAMPLE 4a

A committee of 3 is to be formed from a group of 20 people. How many different

committees are possible?

Solution. There are





20 · 19 · 18

3 · 2 · 1

= 1140 possible committees. .

EXAMPLE 4b

From a group of 5 women and 7 men, how many different committees consisting of

2 women and 3 men can be formed? What if 2 of the men are feuding and refuse to

serve on the committee together?

Solution. As there are





possible groups of 2 women, and





possible

groups of 3 men, it follows from the basic principle that there are









5 · 4

2 · 1



7 · 6 · 5

3 · 2 · 1

= 350 possible committees consisting of 2 women and 3 men.

Now suppose that 2 of the men refuse to serve together. Because a total of







= 5 out of the





= 35 possible groups of 3 men contain both of

the feuding men, it follows that there are 35 − 5 = 30 groups that do not contain

both of the feuding men. Because there are still





= 10 ways to choose the 2

women, there are 30 · 10 = 300 possible committees in this case. .

†

By convention, 0! is deﬁned to be 1. Thus,









= 1. We also take





to be equal

to 0 when either i < 0ori > n.

Section 1.4 Combinations 7

EXAMPLE 4c

Consider a set of n antennas of which m are defective and n − m are functional

and assume that all of the defectives and all of the functionals are considered indis-

tinguishable. How many linear orderings are there in which no two defectives are

consecutive?

Solution. Imagine that the n − m functional antennas are lined up among them-

selves. Now, if no two defectives are to be consecutive, then the spaces between the

functional antennas must each contain at most one defective antenna. That is, in the

n − m + 1 possible positions—represented in Figure 1.1 by carets—between the

n − m functional antennas, we must select m of these in which to put the defective

antennas. Hence, there are



n − m + 1



possible orderings in which there is at

least one functional antenna between any two defective ones. .

1 . . .

1 ⫽ functional

⫽ place for at most one defective

FIGURE 1.1: No consecutive defectives

A useful combinatorial identity is







n − 1

r − 1





n − 1



1 … r … n (4.1)

Equation (4.1) may be proved analytically or by the following combinatorial argu-

ment: Consider a group of n objects, and ﬁx attention on some particular one of these

objects—call it object 1. Now, there are



n − 1

r − 1



groups of size r that contain object

1 (since each such group is formed by selecting r − 1 from the remaining n − 1

objects). Also, there are



n − 1



groups of size r that do not contain object 1. As

there is a total of





groups of size r, Equation (4.1) follows.

The values





are often referred to as binomial coefﬁcients because of their

prominence in the binomial theorem.

The binomial theorem

(x + y)



k=0





n−k

(4.2)

We shall present two proofs of the binomial theorem. The ﬁrst is a proof by math-

ematical induction, and the second is a proof based on combinatorial considerations.

8 Chapter 1 Combinatorial Analysis

Proof of the Binomial Theorem by Induction: When n = 1, Equation (4.2) reduces to

x + y =









= y + x

Assume Equation (4.2) for n − 1. Now,

(x + y)

= (x + y)(x + y)

n−1

= (x + y)

n−1



k=0



n − 1



n−1−k

n−1



k=0



n − 1



k+1

n−1−k

n−1



k=0



n − 1



n−k

Letting i = k + 1intheﬁrstsumandi = k in the second sum, we ﬁnd that

(x + y)



i=1



n − 1

i − 1



n−i

n−1



i=0



n − 1



n−i

= x

n−1



i=1





n − 1

i − 1





n − 1





n−i

+ y

= x

n−i



i=1





n−i

+ y



i=0





n−i

where the next-to-last equality follows by Equation (4.1). By induction, the theorem

is now proved.

Combinatorial Proof of the Binomial Theorem: Consider the product

+ y

)(x

+ y

) ···(x

+ y

)

Its expansion consists of the sum of 2

terms, each term being the product of n factors.

Furthermore, each of the 2

terms in the sum will contain as a factor either x

or y

for each i = 1, 2, ..., n. For example,

+ y

)(x

+ y

) = x

+ x

+ y

Now, how many of the 2

terms in the sum will have k of the x

’s and (n − k) of the y

’s

as factors? As each term consisting of k of the x

’s and (n − k) of the y

’s corresponds

to a choice of a group of k from the n values x

, x

, ..., x

, there are





such terms.

Thus, letting x

= x, y

= y, i = 1, ..., n, we see that

(x + y)



k=0





n−k

Section 1.5 Multinomial Coefﬁcients 9

EXAMPLE 4d

Expand (x + y)

Solution.

(x + y)













y +





= y

+ 3xy

+ 3x

y + x

EXAMPLE 4e

How many subsets are there of a set consisting of n elements?

Solution. Since there are





subsets of size k, the desired answer is



k=0





= (1 + 1)

= 2

This result could also have been obtained by assigning either the number 0 or the

number 1 to each element in the set. To each assignment of numbers, there corre-

sponds, in a one-to-one fashion, a subset, namely, that subset consisting of all ele-

ments that were assigned the value 1. As there are 2

possible assignments, the result

follows.

Note that we have included the set consisting of 0 elements (that is, the null set)

as a subset of the original set. Hence, the number of subsets that contain at least one

element is 2

− 1. .

1.5 MULTINOMIAL COEFFICIENTS

In this section, we consider the following problem: A set of n distinct items is to be

divided into r distinct groups of respective sizes n

, n

, ..., n

, where



i=1

= n.

How many different divisions are possible? To answer this question, we note that

there are





possible choices for the ﬁrst group; for each choice of the ﬁrst group,

there are



n − n



possible choices for the second group; for each choice of the

ﬁrst two groups, there are



n − n

− n



possible choices for the third group; and

so on. It then follows from the generalized version of the basic counting principle that

there are





n − n



···



n − n

− n

− ··· − n

r−1



(n − n

)! n

(n − n

− n

)! n

···

(n − n

− n

− ··· − n

r−1

0! n

! n

! ···n

possible divisions.

10 Chapter 1 Combinatorial Analysis

Another way to see this result is to consider the n values 1, 1, ...,1,2,...,2,...,

r, ..., r, where i appears n

times, for i = 1, ..., r. Every permutation of these values

corresponds to a division of the n items into the r groups in the following manner:

Let the permutation i

, i

, ..., i

correspond to assigning item 1 to group i

, item 2 to

group i

, and so on. For instance, if n = 8andifn

= 4, n

= 3, and n

= 1, then

the permutation 1, 1, 2, 3, 2, 1, 2, 1 corresponds to assigning items 1, 2, 6, 8 to the ﬁrst

group, items 3, 5, 7 to the second group, and item 4 to the third group. Because every

permutation yields a division of the items and every possible division results from

some permutation, it follows that the number of divisions of n items into r distinct

groups of sizes n

, n

, ..., n

is the same as the number of permutations of n items of

which n

are alike, and n

are alike, ...,andn

are alike, which was shown in Section

1.3toequal

! ···n

Notation

If n

+ n

+ ··· + n

= n, we deﬁne



, n

, ..., n





, n

, ..., n



! n

! ···n

Thus,



, n

, ..., n



represents the number of possible divisions of n distinct

objects into r distinct groups of respective sizes n

, n

, ..., n

EXAMPLE 5a

A police department in a small city consists of 10 ofﬁcers. If the department policy is

to have 5 of the ofﬁcers patrolling the streets, 2 of the ofﬁcers working full time at the

station, and 3 of the ofﬁcers on reserve at the station, how many different divisions of

the 10 ofﬁcers into the 3 groups are possible?

Solution. There are

10!

5! 2! 3!

= 2520 possible divisions. .

EXAMPLE 5b

Ten children are to be divided into an A team and a B team of 5 each. The A team

will play in one league and the B team in another. How many different divisions are

possible?

Solution. There are

10!

5! 5!

= 252 possible divisions. .

EXAMPLE 5c

In order to play a game of basketball, 10 children at a playground divide themselves

into two teams of 5 each. How many different divisions are possible?

Section 1.5 Multinomial Coefﬁcients 11

Solution. Note that this example is different from Example 5b because now the order

of the two teams is irrelevant. That is, there is no A and B team, but just a division

consisting of 2 groups of 5 each. Hence, the desired answer is

10!/(5! 5!)

= 126 .

The proof of the following theorem, which generalizes the binomial theorem, is

left as an exercise.

The multinomial theorem

(

+ x

+ ··· + x

)



, ..., n

) :

+ ··· + n

= n



, n

, ..., n



···x

That is, the sum is over all nonnegative integer-valued vectors (n

, n

, ..., n

) such

that n

+ n

+ ··· + n

= n.

The numbers



, n

, ..., n



are known as multinomial coefﬁcients.

EXAMPLE 5d

In the ﬁrst round of a knockout tournament involving n = 2

players, the n players

are divided into n/2 pairs, with each of these pairs then playing a game. The losers

of the games are eliminated while the winners go on to the next round, where the

process is repeated until only a single player remains. Suppose we have a knockout

tournament of 8 players.

(a) How many possible outcomes are there for the initial round? (For instance, one

outcome is that 1 beats 2, 3 beats 4, 5 beats 6, and 7 beats 8. )

(b) How many outcomes of the tournament are possible, where an outcome gives

complete information for all rounds?

Solution. One way to determine the number of possible outcomes for the initial

round is to ﬁrst determine the number of possible pairings for that round. To do

so, note that the number of ways to divide the 8 players into a ﬁrst pair, a second pair,

a third pair, and a fourth pair is



2, 2, 2, 2



. Thus, the number of possible pair-

ings when there is no ordering of the 4 pairs is

. For each such pairing, there are

2 possible choices from each pair as to the winner of that game, showing that there

are

8!2

possible results of round 1. (Another way to see this is to note that

there are





possible choices of the 4 winners and, for each such choice, there are

4! ways to pair the 4 winners with the 4 losers, showing that there are 4!





possible results for the ﬁrst round.)

12 Chapter 1 Combinatorial Analysis

Similarly, for each result of round 1, there are

possible outcomes of round 2,

and for each of the outcomes of the ﬁrst two rounds, there are

possible outcomes

of round 3. Consequently, by the generalized basic principle of counting, there are

= 8! possible outcomes of the tournament. Indeed, the same argument

can be used to show that a knockout tournament of n = 2

players has n! possible

outcomes.

Knowing the preceding result, it is not difﬁcult to come up with a more direct

argument by showing that there is a one-to-one correspondence between the set of

possible tournament results and the set of permutations of 1, ..., n. To obtain such

a correspondence, rank the players as follows for any tournament result: Give the

tournament winner rank 1, and give the ﬁnal-round loser rank 2. For the two play-

ers who lost in the next-to-last round, give rank 3 to the one who lost to the player

ranked 1 and give rank 4 to the one who lost to the player ranked 2. For the four

players who lost in the second-to-last round, give rank 5 to the one who lost to player

ranked 1, rank 6 to the one who lost to the player ranked 2, rank 7 to the one who

lost to the player ranked 3, and rank 8 to the one who lost to the player ranked 4.

Continuing on in this manner gives a rank to each player. (A more succinct descrip-

tion is to give the winner of the tournament rank 1 and let the rank of a player who

lost in a round having 2

matches be 2

plus the rank of the player who beat him, for

k = 0, ..., m − 1.) In this manner, the result of the tournament can be represented

by a permutation i

, i

, ..., i

, where i

is the player who was given rank j. Because

different tournament results give rise to different permutations, and because there is

a tournament result for each permutation, it follows that there are the same number

of possible tournament results as there are permutations of 1, ..., n. .

EXAMPLE 5e

+ x

)



2, 0, 0





0, 2, 0





0, 0, 2





1, 1, 0





1, 0, 1





0, 1, 1



= x

+ x

+ 2x

∗

1.6 THE NUMBER OF INTEGER SOLUTIONS OF EQUATIONS

There are r

possible outcomes when n distinguishable balls are to be distributed into

r distinguishable urns. This result follows because each ball may be distributed into

any of r possible urns. Let us now, however, suppose that the n balls are indistinguish-

able from each other. In this case, how many different outcomes are possible? As the

balls are indistinguishable, it follows that the outcome of the experiment of distribut-

ing the n balls into r urns can be described by a vector (x

, x

, ..., x

), where x

denotes

the number of balls that are distributed into the ith urn. Hence, the problem reduces

to ﬁnding the number of distinct nonnegative integer-valued vectors (x

, x

, ..., x

)

such that

+ x

+ ··· + x

= n

∗

Asterisks denote material that is optional.

Section 1.6 The Number of Integer Solutions of Equations 13

To compute this number, let us start by considering the number of positive integer-

valued solutions. Toward that end, imagine that we have n indistinguishable objects

lined up and that we want to divide them into r nonempty groups. To do so, we can

select r − 1ofthen − 1 spaces between adjacent objects as our dividing points. (See

Figure 1.2.) For instance, if we have n = 8andr = 3 and we choose the 2 divisors so

as to obtain

ooo|ooo|oo

. . .

n objects 0

Choose r ⫺ 1 of the spaces

FIGURE 1.2: Number of positive solutions

then the resulting vector is x

= 3, x

= 2. As there are



n − 1

r − 1



possible

selections, we have the following proposition.

Proposition 6.1. There are



n − 1

r − 1



distinct positive integer-valued vectors (x

, ..., x

) satisfying the equation

+ x

+ ··· + x

= nx

> 0, i = 1, ..., r

To obtain the number of nonnegative (as opposed to positive) solutions, note

that the number of nonnegative solutions of x

+ x

+ ··· + x

= n is the same

as the number of positive solutions of y

+ ··· + y

= n + r (seen by letting

= x

+ 1, i = 1, ..., r). Hence, from Proposition 6.1, we obtain the following

proposition.

Proposition 6.2. There are



n + r − 1

r − 1



distinct nonnegative integer-valued vec-

tors (x

, x

, ..., x

) satisfying the equation

+ x

+ ··· + x

= n (6.1)

EXAMPLE 6a

How many distinct nonnegative integer-valued solutions of x

+ x

= 3 are possible?

Solution. There are



3 + 2 − 1

2 − 1



= 4 such solutions: (0, 3), (1, 2), (2, 1), (3, 0). .

EXAMPLE 6b

An investor has 20 thousand dollars to invest among 4 possible investments. Each

investment must be in units of a thousand dollars. If the total 20 thousand is to be

14 Chapter 1 Combinatorial Analysis

invested, how many different investment strategies are possible? What if not all the

money need be invested?

Solution. If we let x

, i = 1, 2, 3, 4, denote the number of thousands invested in

investment i, then, when all is to be invested, x

, x

are integers satisfying the

equation

+ x

= 20 x

Ú 0

Hence, by Proposition 6.2, there are





= 1771 possible investment strategies. If

not all of the money need be invested, then if we let x

denote the amount kept in

reserve, a strategy is a nonnegative integer-valued vector (x

, x

) satisfying

the equation

+ x

= 20

Hence, by Proposition 6.2, there are now





= 10,626 possible strategies. .

EXAMPLE 6c

How many terms are there in the multinomial expansion of (x

+ x

+ ··· + x

)

Solution.

+ x

+ ··· + x

)





, ..., n



···x

where the sum is over all nonnegative integer-valued (n

, ..., n

) such that n

+···+

= n. Hence, by Proposition 6.2, there are



n + r − 1

r − 1



such terms. .

EXAMPLE 6d

Let us consider again Example 4c, in which we have a set of n items, of which m are

(indistinguishable and) defective and the remaining n − m are (also indistinguishable

and) functional. Our objective is to determine the number of linear orderings in which

no two defectives are next to each other. To determine this number, let us imagine

that the defective items are lined up among themselves and the functional ones are

now to be put in position. Let us denote x

as the number of functional items to the

left of the ﬁrst defective, x

as the number of functional items between the ﬁrst two

defectives, and so on. That is, schematically, we have

0 x

0 ···x

0 x

m+1

Now, there will be at least one functional item between any pair of defectives as long

as x

> 0, i = 2, ..., m. Hence, the number of outcomes satisfying the condition is the

number of vectors x

, ..., x

m+1

that satisfy the equation

+ ··· + x

m+1

= n − mx

Ú 0, x

m+1

Ú 0, x

> 0, i = 2, ..., m

Summary 15

But, on letting y

= x

+ 1, y

= x

, i = 2, ..., m, y

m+1

= x

m+1

+ 1, we see that

this number is equal to the number of positive vectors (y

, ..., y

m+1

) that satisfy the

equation

+ y

+ ··· + y

m+1

= n − m + 2

Hence, by Proposition 6.1, there are



n − m + 1



such outcomes, in agreement

with the results of Example 4c.

Suppose now that we are interested in the number of outcomes in which each pair

of defective items is separated by at least 2 functional items. By the same reason-

ing as that applied previously, this would equal the number of vectors satisfying the

equation

+ ··· + x

m+1

= n − mx

Ú 0, x

m+1

Ú 0, x

Ú 2, i = 2, ..., m

Upon letting y

= x

+ 1, y

= x

− 1, i = 2, ..., m, y

m+1

= x

m+1

+ 1, we see that

this is the same as the number of positive solutions of the equation

+ ··· + y

m+1

= n − 2m + 3

Hence, from Proposition 6.1, there are



n − 2m + 2



such outcomes. .

SUMMARY

The basic principle of counting states that if an experiment consisting of two phases is

such that there are n possible outcomes of phase 1 and, for each of these n outcomes,

there are m possible outcomes of phase 2, then there are nm possible outcomes of the

experiment.

There are n! = n(n − 1) ···3 · 2 · 1 possible linear orderings of n items. The

quantity 0! is deﬁned to equal 1.

Let





(n − i)! i!

when 0 … i … n, and let it equal 0 otherwise. This quantity represents the number

of different subgroups of size i that can be chosen from a set of size n. It is often

called a binomial coefﬁcient because of its prominence in the binomial theorem, which

states that

(x + y)



i=0





n−i

For nonnegative integers n

, ..., n

summing to n,



, n

, ..., n



! ···n

is the number of divisions of n items into r distinct nonoverlapping subgroups of sizes

, n

, ..., n

16 Chapter 1 Combinatorial Analysis

PROBLEMS

1. (a) How many different 7-place license plates are

possible if the ﬁrst 2 places are for letters and

the other 5 for numbers?

(b) Repeat part (a) under the assumption that no

letter or number can be repeated in a single

license plate.

2. How many outcome sequences are possible when a

die is rolled four times, where we say, for instance,

that the outcome is 3, 4, 3, 1 if the ﬁrst roll landed

on 3, the second on 4, the third on 3, and the fourth

on 1?

3. Twenty workers are to be assigned to 20 different

jobs, one to each job. How many different assign-

ments are possible?

4. John, Jim, Jay, and Jack have formed a band con-

sisting of 4 instruments. If each of the boys can play

all 4 instruments, how many different arrange-

ments are possible? What if John and Jim can play

all 4 instruments, but Jay and Jack can each play

only piano and drums?

5. For years, telephone area codes in the United

States and Canada consisted of a sequence of three

digits. The ﬁrst digit was an integer between 2 and

9, the second digit was either 0 or 1, and the third

digit was any integer from 1 to 9. How many area

codes were possible? How many area codes start-

ing with a 4 were possible?

6. A well-known nursery rhyme starts as follows:

“As I was going to St. Ives

I met a man with 7 wives.

Each wife had 7 sacks.

Each sack had 7 cats.

Each cat had 7 kittens...”

How many kittens did the traveler meet?

7. (a) In how many ways can 3 boys and 3 girls sit in

arow?

(b) In how many ways can 3 boys and 3 girls sit in

a row if the boys and the girls are each to sit

together?

(d) In how many ways if no two people of the

same sex are allowed to sit together?

8. How many different letter arrangements can be

made from the letters

(a) Fluke?

(b) Propose?

(d) Arrange?

9. A child has 12 blocks, of which 6 are black, 4 are

red, 1 is white, and 1 is blue. If the child puts the

blocks in a line, how many arrangements are pos-

sible?

10. In how many ways can 8 people be seated in a

row if

(a) there are no restrictions on the seating

arrangement?

(b) persons A and B must sit next to each other?

or 2 women can sit next to each other?

(d) there are 5 men and they must sit next to each

other?

(e) there are 4 married couples and each couple

must sit together?

11. In how many ways can 3 novels, 2 mathematics

books, and 1 chemistry book be arranged on a

bookshelf if

(a) the books can be arranged in any order?

(b) the mathematics books must be together and

the novels must be together?

books can be arranged in any order?

12. Five separate awards (best scholarship, best lead-

ership qualities, and so on) are to be presented to

selected students from a class of 30. How many dif-

ferent outcomes are possible if

(a) a student can receive any number of awards?

(b) each student can receive at most 1 award?

13. Consider a group of 20 people. If everyone shakes

hands with everyone else, how many handshakes

take place?

14. How many 5-card poker hands are there?

15. A dance class consists of 22 students, of which 10

are women and 12 are men. If 5 men and 5 women

are to be chosen and then paired off, how many

results are possible?

16. A student has to sell 2 books from a collection of

6 math, 7 science, and 4 economics books. How

many choices are possible if

(a) both books are to be on the same subject?

(b) the books are to be on different subjects?

17. Seven different gifts are to be distributed among

10 children. How many distinct results are possible

if no child is to receive more than one gift?

18. A committee of 7, consisting of 2 Republicans,

2 Democrats, and 3 Independents, is to be cho-

sen from a group of 5 Republicans, 6 Democrats,

and 4 Independents. How many committees are

possible?

19. From a group of 8 women and 6 men, a committee

consisting of 3 men and 3 women is to be formed.

How many different committees are possible if

(a) 2 of the men refuse to serve together?

(b) 2 of the women refuse to serve together?

Problems 17

20. A person has 8 friends, of whom 5 will be invited

to a party.

(a) How many choices are there if 2 of the friends

are feuding and will not attend together?

(b) How many choices if 2 of the friends will only

attend together?

21. Consider the grid of points shown here. Suppose

that, starting at the point labeled A, you can go one

step up or one step to the right at each move. This

procedure is continued until the point labeled B is

reached. How many different paths from A to B

are possible?

Hint: Note that to reach B from A, you must take

4 steps to the right and 3 steps upward.

22. In Problem 21, how many different paths are there

from A to B that go through the point circled in

the following lattice?

23. A psychology laboratory conducting dream

research contains 3 rooms, with 2 beds in each

room. If 3 sets of identical twins are to be assigned

to these 6 beds so that each set of twins sleeps

in different beds in the same room, how many

assignments are possible?

24. Expand (3x

+ y)

25. The game of bridge is played by 4 players, each of

whom is dealt 13 cards. How many bridge deals are

possible?

26. Expand (x

+ 2x

+ 3x

)

27. If 12 people are to be divided into 3 committees of

respective sizes 3, 4, and 5, how many divisions are

possible?

28. If 8 new teachers are to be divided among 4

schools, how many divisions are possible? What if

each school must receive 2 teachers?

29. Ten weight lifters are competing in a team weight-

lifting contest. Of the lifters, 3 are from the United

States, 4 are from Russia, 2 are from China, and 1

is from Canada. If the scoring takes account of the

countries that the lifters represent, but not their

individual identities, how many different outcomes

are possible from the point of view of scores? How

many different outcomes correspond to results in

which the United States has 1 competitor in the

top three and 2 in the bottom three?

30. Delegates from 10 countries, including Russia,

France, England, and the United States, are to

be seated in a row. How many different seat-

ing arrangements are possible if the French and

English delegates are to be seated next to each

other and the Russian and U.S. delegates are not

to be next to each other?

∗

31. If 8 identical blackboards are to be divided among

4 schools, how many divisions are possible? How

many if each school must receive at least 1 black-

board?

∗

32. An elevator starts at the basement with 8 peo-

ple (not including the elevator operator) and dis-

charges them all by the time it reaches the top

ﬂoor, number 6. In how many ways could the oper-

ator have perceived the people leaving the eleva-

tor if all people look alike to him? What if the 8

people consisted of 5 men and 3 women and the

operator could tell a man from a woman?

∗

33. We have 20 thousand dollars that must be invested

among 4 possible opportunities. Each investment

must be integral in units of 1 thousand dollars,

and there are minimal investments that need to be

made if one is to invest in these opportunities. The

minimal investments are 2, 2, 3, and 4 thousand

dollars. How many different investment strategies

are available if

(a) an investment must be made in each opportu-

nity?

(b) investments must be made in at least 3 of the

4 opportunities?

18 Chapter 1 Combinatorial Analysis

THEORETICAL EXERCISES

1. Prove the generalized version of the basic counting

principle.

2. Two experiments are to be performed. The ﬁrst

can result in any one of m possible outcomes. If

the ﬁrst experiment results in outcome i, then the

second experiment can result in any of n

possible

outcomes, i = 1, 2, ..., m. What is the number of

possible outcomes of the two experiments?

3. In how many ways can r objects be selected from a

set of n objects if the order of selection is consid-

ered relevant?

4. There are





different linear arrangements of n

balls of which r are black and n − r are white. Give

a combinatorial explanation of this fact.

5. Determine the number of vectors (x

, ..., x

),such

that each x

is either 0 or 1 and



i=1

Ú k

6. How many vectors x

, ..., x

are there for which

each x

is a positive integer such that 1 … x

… n

and x

< x

< ··· < x

7. Give an analytic proof of Equation (4.1).

8. Prove that



n + m













r − 1



+··· +







Hint: Consider a group of n men and m women.

How many groups of size r are possible?

9. Use Theoretical Exercise 8 to prove that







k=0





10. From a group of n people, suppose that we want to

choose a committee of k, k … n, one of whom is to

be designated as chairperson.

(a) By focusing ﬁrst on the choice of the commit-

tee and then on the choice of the chair, argue

that there are





k possible choices.

(b) By focusing ﬁrst on the choice of the

nonchair committee members and then on

the choice of the chair, argue that there are



k − 1



(n − k + 1) possible choices.

and then on the choice of the other committee

members, argue that there are n



n − 1

k − 1



possible choices.

(d) Conclude from parts (a), (b), and (c) that





= (n − k + 1)



k − 1



= n



n − 1

k − 1



(e) Use the factorial deﬁnition of





to verify

the identity in part (d).

11. The following identity is known as Fermat’s com-

binatorial identity:







i=k



i − 1

k − 1



n Ú k

Give a combinatorial argument (no computations

are needed) to establish this identity.

Hint: Consider the set of numbers 1 through n.

How many subsets of size k have i as their highest-

numbered member?

12. Consider the following combinatorial identity:



k=1





= n · 2

n−1

(a) Present a combinatorial argument for this

identity by considering a set of n people and

determining, in two ways, the number of pos-

sible selections of a committee of any size and

a chairperson for the committee.

Hint:

(i) How many possible selections are there

ofacommitteeofsizek and its chairper-

son?

(ii) How many possible selections are there

of a chairperson and the other commit-

tee members?

(b) Verify the following identity for n =

1, 2, 3, 4, 5:



k=1





= 2

n−2

n(n + 1)

Theoretical Exercises 19

For a combinatorial proof of the preceding,

consider a set of n people and argue that both

sides of the identity represent the number of

different selections of a committee, its chair-

person, and its secretary (possibly the same as

the chairperson).

Hint:

(i) How many different selections result in

the committee containing exactly k peo-

ple?

(ii) How many different selections are there

in which the chairperson and the secre-

tary are the same? (

ANSWER: n2

n−1

(iii) How many different selections result in

the chairperson and the secretary being

different?



k=1





= 2

n−3

(n + 3)

13. Show that, for n > 0,



i=0

(−1)





= 0

Hint: Use the binomial theorem.

14. From a set of n people, a committee of size j is to be

chosen, and from this committee, a subcommittee

of size i, i … j, is also to be chosen.

(a) Derive a combinatorial identity by comput-

ing, in two ways, the number of possible

choices of the committee and subcommittee—

ﬁrst by supposing that the committee is

chosen ﬁrst and then the subcommittee is

chosen, and second by supposing that the

subcommittee is chosen ﬁrst and then the

remaining members of the committee are

chosen.

(b) Use part (a) to prove the following combina-

torial identity:



j=i











n−i

i … n

show that



j=i







(−1)

n−j

= 0 i < n

15. Let H

(n) be the number of vectors x

, ..., x

for

which each x

is a positive integer satisfying 1 …

… n and x

… x

… ··· … x

(a) Without any computations, argue that

(n) = n

(n) =



j=1

k−1

(j) k > 1

Hint: How many vectors are there in which

= j?

(b) Use the preceding recursion to compute

(5).

Hint: First compute H

(n) for n = 1, 2, 3, 4, 5.

16. Consider a tournament of n contestants in which

the outcome is an ordering of these contestants,

with ties allowed. That is, the outcome partitions

the players into groups, with the ﬁrst group consist-

ing of the players that tied for ﬁrst place, the next

group being those that tied for the next-best posi-

tion, and so on. Let N(n) denote the number of dif-

ferent possible outcomes. For instance, N(2) = 3,

since, in a tournament with 2 contestants, player 1

could be uniquely ﬁrst, player 2 could be uniquely

ﬁrst, or they could tie for ﬁrst.

(a) List all the possible outcomes when n = 3.

(b) With N(0) deﬁned to equal 1, argue, without

any computations, that

N(n) =



i=1





N(n − i)

Hint: How many outcomes are there in

which i players tie for last place?

to the following:

N(n) =

n−1



i=0





N(i)

(d) Use the recursion to ﬁnd N(3) and N(4).

17. Present a combinatorial explanation of why







r, n − r



20 Chapter 1 Combinatorial Analysis

18. Argue that



, n

, ..., n





n − 1

− 1, n

, ..., n





n − 1

, n

− 1, ..., n



+ ···



n − 1

, n

, ..., n

− 1



Hint: Use an argument similar to the one used to

establish Equation (4.1).

19. Prove the multinomial theorem.

∗

20. In how many ways can n identical balls be dis-

tributed into r urns so that the ith urn contains at

least m

balls, for each i = 1, ..., r? Assume that

n Ú



i=1

∗

21. Argue that there are exactly





n − 1

n − r + k



solutions of

+ x

+ ··· + x

= n

for which exactly k of the x

are equal to 0.

∗

22. Consider a function f (x

, ..., x

) of n variables.

How many different partial derivatives of order r

does f possess?

∗

23. Determine the number of vectors (x

, ..., x

) such

that each x

is a nonnegative integer and



i=1

… k

SELF-TEST PROBLEMS AND EXERCISES

1. How many different linear arrangements are there

of the letters A, B, C, D, E, F for which

(a) A and B are next to each other?

(b) AisbeforeB?

(d) A is before B and C is before D?

(e) A and B are next to each other and C and D

are also next to each other?

(f) E is not last in line?

2. If 4 Americans, 3 French people, and 3 British

people are to be seated in a row, how many seat-

ing arrangements are possible when people of the

same nationality must sit next to each other?

3. A president, treasurer, and secretary, all different,

are to be chosen from a club consisting of 10 peo-

ple. How many different choices of ofﬁcers are

possible if

(a) there are no restrictions?

(b) A and B will not serve together?

(d) E must be an ofﬁcer?

(e) F will serve only if he is president?

4. A student is to answer 7 out of 10 questions in

an examination. How many choices has she? How

many if she must answer at least 3 of the ﬁrst 5

questions?

5. In how many ways can a man divide 7 gifts among

his 3 children if the eldest is to receive 3 gifts and

the others 2 each?

6. How many different 7-place license plates are pos-

sible when 3 of the entries are letters and 4 are

digits? Assume that repetition of letters and num-

bers is allowed and that there is no restriction on

where the letters or numbers can be placed.

7. Give a combinatorial explanation of the identity







n − r



8. Consider n-digit numbers where each digit is one

of the 10 integers 0, 1, ..., 9. How many such num-

bers are there for which

(a) no two consecutive digits are equal?

(b) 0 appears as a digit a total of i times, i =

0, ..., n?

9. Consider three classes, each consisting of n stu-

dents. From this group of 3n students, a group of 3

students is to be chosen.

(a)

How many choices are possible?

(b)

How many choices are there in which all 3 stu-

dents are in the same class?

3 students are in the same class and the other

student is in a different class?

(d) How many choices are there in which all 3 stu-

dents are in different classes?

(e) Using the results of parts (a) through (d),

write a combinatorial identity.

10. How many 5-digit numbers can be formed from

the integers 1, 2, ..., 9 if no digit can appear more

than twice? (For instance, 41434 is not allowed.)

11. From 10 married couples, we want to select a

group of 6 people that is not allowed to contain

a married couple.

(a) How many choices are there?

(b) How many choices are there if the group must

also consist of 3 men and 3 women?

Self-Test Problems and Exercises 21

12. A committee of 6 people is to be chosen from a

group consisting of 7 men and 8 women. If the

committee must consist of at least 3 women and

at least 2 men, how many different committees are

possible?

∗

13. An art collection on auction consisted of 4 Dalis, 5

van Goghs, and 6 Picassos. At the auction were 5

art collectors. If a reporter noted only the number

of Dalis, van Goghs, and Picassos acquired by each

collector, how many different results could have

been recorded if all of the works were sold?

∗

14. Determine the number of vectors (x

, ..., x

) such

that each x

is a positive integer and



i=1

… k

where k Ú n.

15. A total of n students are enrolled in a review

course for the actuarial examination in probability.

The posted results of the examination will list the

names of those who passed, in decreasing order of

their scores. For instance, the posted result will be

“Brown, Cho” if Brown and Cho are the only ones

to pass, with Brown receiving the higher score.

Assuming that all scores are distinct (no ties), how

many posted results are possible?

16. Howmanysubsetsofsize4ofthesetS =

{1, 2, ...,20} contain at least one of the elements

1, 2, 3, 4, 5?

17. Give an analytic veriﬁcation of









+ k(n − k) +



n − k



,1… k … n

Now, give a combinatorial argument for this

identity.

18. In a certain community, there are 3 families con-

sisting of a single parent and 1 child, 3 families

consisting of a single parent and 2 children, 5 fam-

ilies consisting of 2 parents and a single child, 7

families consisting of 2 parents and 2 children, and

6 families consisting of 2 parents and 3 children. If

a parent and child from the same family are to be

chosen, how many possible choices are there?

19. If there are no restrictions on where the digits and

letters are placed, how many 8-place license plates

consisting of 5 letters and 3 digits are possible if no

repetitions of letters or digits are allowed. What if

the 3 digits must be consecutive?

CHAPTER 2

Axioms of Probability

2.1 INTRODUCTION

2.2 SAMPLE SPACE AND EVENTS

2.3 AXIOMS OF PROBABILITY

2.4 SOME SIMPLE PROPOSITIONS

2.5 SAMPLE SPACES HAVING EQUALLY LIKELY OUTCOMES

2.6 PROBABILITY AS A CONTINUOUS SET FUNCTION

2.7 PROBABILITY AS A MEASURE OF BELIEF

2.1 INTRODUCTION

In this chapter, we introduce the concept of the probability of an event and then show

how probabilities can be computed in certain situations. As a preliminary, however,

we need the concept of the sample space and the events of an experiment.

2.2 SAMPLE SPACE AND EVENTS

Consider an experiment whose outcome is not predictable with certainty. However,

although the outcome of the experiment will not be known in advance, let us suppose

that the set of all possible outcomes is known. This set of all possible outcomes of

an experiment is known as the sample space of the experiment and is denoted by S.

Following are some examples:

1. If the outcome of an experiment consists in the determination of the sex of a

newborn child, then

S ={g, b}

where the outcome g means that the child is a girl and b that it is a boy.

2. If the outcome of an experiment is the order of ﬁnish in a race among the 7

horses having post positions 1, 2, 3, 4, 5, 6, and 7, then

S ={all 7! permutations of (1, 2, 3, 4, 5, 6, 7)}

The outcome (2, 3, 1, 6, 5, 4, 7) means, for instance, that the number 2 horse

comes in ﬁrst, then the number 3 horse, then the number 1 horse, and so on.

3. If the experiment consists of ﬂipping two coins, then the sample space consists

of the following four points:

S ={(H, H), (H, T), (T, H), (T, T)}

The outcome will be (H, H) if both coins are heads, (H, T) if the ﬁrst coin is

heads and the second tails, (T, H) if the ﬁrst is tails and the second heads, and

(T, T) if both coins are tails.

Section 2.2 Sample Space and Events 23

4. If the experiment consists of tossing two dice, then the sample space consists of

the 36 points

S ={(i, j): i, j = 1, 2, 3, 4, 5, 6}

where the outcome (i, j) is said to occur if i appears on the leftmost die and j on

the other die.

5. If the experiment consists of measuring (in hours) the lifetime of a transistor,

then the sample space consists of all nonnegative real numbers; that is,

S ={x:0 … x < q}

Any subset E of the sample space is known as an event. In other words, an event is

a set consisting of possible outcomes of the experiment. If the outcome of the experi-

ment is contained in E, then we say that E has occurred. Following are some examples

of events.

In the preceding Example 1, if E ={g},thenE is the event that the child is a girl.

Similarly, if F ={b},thenF is the event that the child is a boy.

In Example 2, if

E ={all outcomes in S starting with a 3}

then E is the event that horse 3 wins the race.

In Example 3, if E ={(H, H), (H, T)},thenE is the event that a head appears on

the ﬁrst coin.

In Example 4, if E ={(1, 6),

(2, 5), (3, 4), (4, 3), (5, 2), (6, 1)},thenE is the event that

the sum of the dice equals 7.

In Example 5, if E ={x:0 … x … 5},thenE is the event that the transistor does

not last longer than 5 hours.

For any two events E and F of a sample space S, we deﬁne the new event E ∪ F

to consist of all outcomes that are either in E or in F or in both E and F. That is, the

event E ∪ F will occur if either E or F occurs. For instance, in Example 1, if event

E ={g} and F ={b},then

E ∪ F ={g, b}

That is, E ∪ F is the whole sample space S. In Example 3, if E ={(H, H), (H

, T)} and

F ={(T, H)},then

E ∪ F ={(H, H), (H, T), (T, H)}

Thus, E ∪ F would occur if a head appeared on either coin.

The event E ∪ F is called the union of the event E and the event F.

Similarly, for any two events E and F, we may also deﬁne the new event EF, called

the intersection of E and F, to consist of all outcomes that are both in E and in F.

That is, the event EF (sometimes written E ∩ F) will occur only if both E and F

occur. For instance, in Example 3, if E ={(H, H), (H, T), (T, H)}

is the event that at

least 1 head occurs and F ={(H, T), (T, H), (T, T)} is the event that at least 1 tail

occurs, then

EF ={(H, T), (T, H)}

is the event that exactly 1 head and 1 tail occur. In example 4, if E ={(1, 6), (2, 5),

(3, 4), (4, 3), (5, 2), (6, 1)} is the event that the sum of the dice is 7 and F ={(1, 5), (2, 4),

(3, 3), (4, 2), (5, 1)} is the event that the sum is 6, then the event EF does not contain

24 Chapter 2 Axioms of Probability

any outcomes and hence could not occur. To give such an event a name, we shall refer

to it as the null event and denote it by Ø. (That is, Ø refers to the event consisting of

no outcomes.) If EF = Ø, then E and F are said to be mutually exclusive.

We deﬁne unions and intersections of more than two events in a similar manner.

If E

, E

, ... are events, then the union of these events, denoted by



n=1

, is deﬁned

to be that event which consists of all outcomes that are in E

for at least one value

of n = 1, 2, .... Similarly, the intersection of the events E

, denoted by



n=1

,is

deﬁned to be the event consisting of those outcomes which are in all of the events

, n = 1, 2, ....

Finally, for any event E, we deﬁne the new event E

, referred to as the com-

plement of E, to consist of all outcomes in the sample space S that are not in E.

That is, E

will occur if and only if E does not occur. In Example 4, if event E =

{(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)},thenE

will occur when the sum of the dice

does not equal 7. Note that because the experiment must result in some outcome, it

follows that S

= Ø.

For any two events E and F, if all of the outcomes in E are also in F, then we say

that E is contained in F,orE is a subset of F, and write E ( F (or equivalently, F ) E,

which we sometimes say as F is a superset of E). Thus, if E ( F, then the occurrence

of E implies the occurrence of F.IfE ( F and F ( E, we say that E and F are equal

and write E = F.

A graphical representation that is useful for illustrating logical relations among

events is the Venn diagram. The sample space S is represented as consisting of all

the outcomes in a large rectangle, and the events E, F, G, ... are represented as con-

sisting of all the outcomes in given circles within the rectangle. Events of interest

can then be indicated by shading appropriate regions of the diagram. For instance, in

the three Venn diagrams shown in Figure 2.1, the shaded areas represent, respec-

tively, the events E ∪ F, EF,andE

. The Venn diagram in Figure 2.2 indicates

that E ( F.

EFEF

(a) Shaded region: E 傼 F. (b) Shaded region: EF.

FIGURE 2.1: Venn Diagrams

Section 2.2 Sample Space and Events 25

FIGURE 2.2: E ( F

The operations of forming unions, intersections, and complements of events obey

certain rules similar to the rules of algebra. We list a few of these rules:

Commutative laws E ∪F = F ∪EEF= FE

Associative laws (E ∪F) ∪G = E ∪(F ∪G)(EF)G = E(FG)

Distributive laws (E ∪F)G = EG ∪FG EF ∪G = (E ∪G)(F ∪G)

These relations are veriﬁed by showing that any outcome that is contained in the

event on the left side of the equality sign is also contained in the event on the right

side, and vice versa. One way of showing this is by means of Venn diagrams. For

instance, the distributive law may be veriﬁed by the sequence of diagrams in

Figure 2.3.

(a) Shaded region: EG.

(b) Shaded region: FG.

FIGURE 2.3: (E ∪F)G = EG ∪ FG

26 Chapter 2 Axioms of Probability

The following useful relationships between the three basic operations of forming

unions, intersections, and complements are known as DeMorgan’s laws:

⎛

⎝



i=1

⎞

⎠



i=1

⎛

⎝



i=1

⎞

⎠



i=1

To prove DeMorgan’s laws, suppose ﬁrst that x is an outcome of





i=1



. Then

x is not contained in



i=1

, which means that x is not contained in any of the events

, i = 1, 2, ..., n, implying that x is contained in E

for all i = 1, 2, ..., n and thus is

contained in



i=1

. To go the other way, suppose that x is an outcome of



i=1

. Then

x is contained in E

for all i = 1, 2, ..., n, which means that x is not contained in E

for

any i = 1, 2, ..., n, implying that x is not contained in



, in turn implying that x is

contained in







. This proves the ﬁrst of DeMorgan’s laws.

To prove the second of DeMorgan’s laws, we use the ﬁrst law to obtain

⎛

⎝



i=1

⎞

⎠



i=1

)

which, since (E

)

= E, is equivalent to

⎛

⎝



⎞

⎠



Taking complements of both sides of the preceding equation yields the result we seek,

namely,



⎛

⎝



⎞

⎠

2.3 AXIOMS OF PROBABILITY

One way of deﬁning the probability of an event is in terms of its relative frequency.

Such a deﬁnition usually goes as follows: We suppose that an experiment, whose sam-

ple space is S, is repeatedly performed under exactly the same conditions. For each

event E of the sample space S, we deﬁne n(E) to be the number of times in the ﬁrst n

repetitions of the experiment that the event E occurs. Then P(E), the probability of

the event E, is deﬁned as

P(E) = lim

n→q

n(E)

Section 2.3 Axioms of Probability 27

That is, P(E) is deﬁned as the (limiting) proportion of time that E occurs. It is thus

the limiting frequency of E.

Although the preceding deﬁnition is certainly intuitively pleasing and should always

be kept in mind by the reader, it possesses a serious drawback: How do we know that

n(E)/n will converge to some constant limiting value that will be the same for each

possible sequence of repetitions of the experiment? For example, suppose that the

experiment to be repeatedly performed consists of ﬂipping a coin. How do we know

that the proportion of heads obtained in the ﬁrst n ﬂips will converge to some value

as n gets large? Also, even if it does converge to some value, how do we know that,

if the experiment is repeatedly performed a second time, we shall obtain the same

limiting proportion of heads?

Proponents of the relative frequency deﬁnition of probability usually answer this

objection by stating that the convergence of n(E)/n to a constant limiting value is an

assumption, or an axiom, of the system. However, to assume that n(E)/n will neces-

sarily converge to some constant value seems to be an extraordinarily complicated

assumption. For, although we might indeed hope that such a constant limiting fre-

quency exists, it does not at all seem to be a priori evident that this need be the

case. In fact, would it not be more reasonable to assume a set of simpler and more

self-evident axioms about probability and then attempt to prove that such a con-

stant limiting frequency does in some sense exist? The latter approach is the modern

axiomatic approach to probability theory that we shall adopt in this text. In particular,

we shall assume that, for each event E in the sample space S, there exists a value P (E ),

referred to as the probability of E. We shall then assume that all these probabilities

satisfy a certain set of axioms, which, we hope the reader will agree, is in accordance

with our intuitive notion of probability.

Consider an experiment whose sample space is S. For each event E of the sample

space S, we assume that a number P(E) is deﬁned and satisﬁes the following three

axioms:

Axiom 1

0 … P (E) … 1

Axiom 2

P(S) = 1

Axiom 3

For any sequence of mutually exclusive events E

, E

, ... (that is, events for which

= Ø when i Z j),

⎛

⎝



i=1

⎞

⎠



i=1

P(E

)

We refer to P(E) as the probability of the event E.

Thus, Axiom 1 states that the probability that the outcome of the experiment is an

outcome in E is some number between 0 and 1. Axiom 2 states that, with probability

1, the outcome will be a point in the sample space S. Axiom 3 states that, for any

sequence of mutually exclusive events, the probability of at least one of these events

occurring is just the sum of their respective probabilities.

28 Chapter 2 Axioms of Probability

If we consider a sequence of events E

, E

, ..., where E

= S and E

= Øfori > 1,

then, because the events are mutually exclusive and because S =



i=1

, we have, from

Axiom 3,

P(S) =



i=1

P(E

) = P(S) +



i=2

P(Ø)

implying that

P(Ø) = 0

That is, the null event has probability 0 of occurring.

Note that it follows that, for any ﬁnite sequence of mutually exclusive events E

, ..., E

⎛

⎝



⎞

⎠



i=1

P(E

) (3.1)

This equation follows from Axiom 3 by deﬁning E

as the null event for all values

of i greater than n. Axiom 3 is equivalent to Equation (3.1) when the sample space

is ﬁnite. (Why?) However, the added generality of Axiom 3 is necessary when the

sample space consists of an inﬁnite number of points.

EXAMPLE 3a

If our experiment consists of tossing a coin and if we assume that a head is as likely

to appear as a tail, then we would have

P({H}) = P({T}) =

On the other hand, if the coin were biased and we felt that a head were twice as likely

to appear as a tail, then we would have

P({H}) =

P({T}) =

EXAMPLE 3b

If a die is rolled and we suppose that all six sides are equally likely to appear, then

we would have P({1}) = P({2}) = P({3}) = P({4}) = P({5}) = P({6}) =

. From

Axiom 3, it would thus follow that the probability of rolling an even number would

equal

P({2, 4, 6}) = P({2}) + P({4}) + P({6}) =

The assumption of the existence of a set function P, deﬁned on the events of a

sample space S and satisfying Axioms 1, 2, and 3, constitutes the modern mathemat-

ical approach to probability theory. Hopefully, the reader will agree that the axioms

are natural and in accordance with our intuitive concept of probability as related to

chance and randomness. Furthermore, using these axioms we shall be able to prove

that if an experiment is repeated over and over again, then, with probability 1, the

proportion of time during which any speciﬁc event E occurs will equal P(E). This

result, known as the strong law of large numbers, is presented in Chapter 8. In addi-

tion, we present another possible interpretation of probability—as being a measure

of belief—in Section 2.7.

Section 2.4 Some Simple Propositions 29

Technical Remark. We have supposed that P(E) is deﬁned for all the events E

of the sample space. Actually, when the sample space is an uncountably inﬁnite set,

P(E) is deﬁned only for a class of events called measurable. However, this restriction

need not concern us, as all events of any practical interest are measurable.

2.4 SOME SIMPLE PROPOSITIONS

In this section, we prove some simple propositions regarding probabilities. We ﬁrst

note that, since E and E

are always mutually exclusive and since E ∪ E

= S,we

have, by Axioms 2 and 3,

1 = P(S) = P(E ∪ E

) = P(E) + P(E

)

Or, equivalently, we have Proposition 4.1.

Proposition 4.1.

P(E

) = 1 − P(E)

In words, Proposition 4.1 states that the probability that an event does not occur is

1 minus the probability that it does occur. For instance, if the probability of obtaining

a head on the toss of a coin is

, then the probability of obtaining a tail must be

Our second proposition states that if the event E is contained in the event F,then

the probability of E is no greater than the probability of F.

Proposition 4.2. If E ( F,thenP(E) … P(F).

Proof. Since E ( F, it follows that we can express F as

F = E ∪ E

Hence, because E and E

F are mutually exclusive, we obtain, from Axiom 3,

P(F) = P(E) + P(E

which proves the result, since P(E

F) Ú 0.

Proposition 4.2 tells us, for instance, that the probability of rolling a 1 with a die is

less than or equal to the probability of rolling an odd value with the die.

The next proposition gives the relationship between the probability of the union

of two events, expressed in terms of the individual probabilities, and the probability

of the intersection of the events.

Proposition 4.3.

P(E ∪ F) = P(E) + P(F) − P(EF)

Proof. To derive a formula for P(E ∪ F), we ﬁrst note that E ∪ F can be written

as the union of the two disjoint events E and E

F. Thus, from Axiom 3, we obtain

P(E ∪ F) = P(E ∪ E

= P(E) + P (E

Furthermore, since F = EF ∪ E

F, we again obtain from Axiom 3

P(F) = P(EF) + P(E

30 Chapter 2 Axioms of Probability

FIGURE 2.4: Venn Diagram

I IIIII

FIGURE 2.5: Venn Diagram in Sections

or, equivalently,

P(E

F) = P(F) − P(EF)

thereby completing the proof.

Proposition 4.3 could also have been proved by making use of the Venn diagram

in Figure 2.4.

Let us divide E ∪ F into three mutually exclusive sections, as shown in Figure 2.5.

In words, section I represents all the points in E that are not in F (that is, EF

section II represents all points both in E and in F (that is, EF), and section III repre-

sents all points in F that are not in E (that is, E

F).

From Figure 2.5, we see that

E ∪ F = I ∪ II ∪ III

E = I ∪ II

F = II ∪ III

As I, II, and III are mutually exclusive, it follows from Axiom 3 that

P(E ∪ F) = P(I) + P (II) + P (III)

P(E) = P(I) + P(II)

P(F) = P(II) + P(III)

which shows that

P(E ∪ F) = P(E) + P(F) − P(II)

and Proposition 4.3 is proved, since II = EF.

EXAMPLE 4a

J is taking two books along on her holiday vacation. With probability .5, she will like

the ﬁrst book; with probability .4, she will like the second book; and with probabil-

ity .3, she will like both books. What is the probability that she likes neither book?

Section 2.4 Some Simple Propositions 31

Solution. Let B

denote the event that J likes book i, i = 1, 2. Then the probability

that she likes at least one of the books is

P(B

∪ B

) = P(B

) + P (B

) − P (B

) = .5 + .4 − .3 = .6

Because the event that J likes neither book is the complement of the event that she

likes at least one of them, we obtain the result

P(B

) = P



∪ B

)



= 1 − P(B

∪ B

) = .4 .

We may also calculate the probability that any one of the three events E, F,andG

occurs, namely,

P(E ∪ F ∪ G) = P[(E ∪ F) ∪ G]

which, by Proposition 4.3, equals

P(E ∪ F) + P(G) − P[(E ∪ F)G]

Now, it follows from the distributive law that the events (E ∪ F)G and EG ∪ FG are

equivalent; hence, from the preceding equations, we obtain

P(E ∪ F ∪ G)

= P(E) + P (F) − P(EF) + P(G) − P(EG ∪ FG)

= P(E) + P (F) − P(EF) + P(G) − P(EG) − P(FG) + P(EGFG)

= P(E) + P (F) + P(G) − P(EF) − P(EG) − P(FG) + P(EFG)

In fact, the following proposition, known as the inclusion–exclusion identity, can

be proved by mathematical induction:

Proposition 4.4.

P(E

∪ E

∪ ··· ∪ E

) =



i=1

P(E

) −



P(E

) + ···

+ (−1)

r+1



<···<i

P(E

···E

)

+ ··· + (−1)

n+1

P(E

···E

)

The summation



<···<i

P(E

···E

) is taken over all of the





possible sub-

sets of size r of the set {1, 2, ..., n}.

In words, Proposition 4.4 states that the probability of the union of n events equals

the sum of the probabilities of these events taken one at a time, minus the sum of the

probabilities of these events taken two at a time, plus the sum of the probabilities of

these events taken three at a time, and so on.

Remarks. 1. For a noninductive argument for Proposition 4.4, note ﬁrst that if an

outcome of the sample space is not a member of any of the sets E

, then its probability

does not contribute anything to either side of the equality. Now, suppose that an

outcome is in exactly m of the events E

, where m > 0. Then, since it is in



,its

32 Chapter 2 Axioms of Probability

probability is counted once in P







; also, as this outcome is contained in





subsets of the type E

···E

, its probability is counted





−









− ··· ;





times on the right of the equality sign in Proposition 4.4. Thus, for m > 0, we must

show that

1 =





−









− ··· ;





However, since 1 =





, the preceding equation is equivalent to



i=0





(−1)

= 0

and the latter equation follows from the binomial theorem, since

0 = (−1 + 1)



i=0





(−1)

(1)

m−i

2. The following is a succinct way of writing the inclusion–exclusion identity:

P(∪

i=1

) =



r=1

(−1)

r+1



<···<i

P(E

···E

)

3. In the inclusion–exclusion identity, going out one term results in an upper bound

on the probability of the union, going out two terms results in a lower bound on the

probability, going out three terms results in an upper bound on the probability, going

out four terms results in a lower bound, and so on. That is, for events E

, ..., E

we have

P(∪

i=1

) …



i=1

P(E

) (4.1)

P(∪

i=1

) Ú



i=1

P(E

) −



j<i

P(E

) (4.2)

P(∪

i=1

) …



i=1

P(E

) −



j<i

P(E

) +



k<j<i

P(E

) (4.3)

and so on. To prove the validity of these bounds, note the identity

∪

i=1

= E

∪ E

∪ ··· ∪ E

···E

n−1

Section 2.5 Sample Spaces Having Equally Likely Outcomes 33

That is, at least one of the events E

occurs if E

occurs, or if E

does not occur but

does, or if E

and E

do not occur but E

does, and so on. Because the right-hand

side is the union of disjoint events, we obtain

P(∪

i=1

) = P(E

) + P (E

) + ... + P(E

···E

n−1

)

= P(E

) +



i=2

P(E

···E

i−1

) (4.4)

Now, let B

= E

···E

i−1

= (∪

j<i

)

be the event that none of the ﬁrst i − 1 events

occur. Applying the identity

P(E

) = P(B

) + P (B

)

shows that

P(E

) = P(E

···E

i−1

) + P (E

∪

j<i

)

or, equivalently,

P(E

···E

i−1

) = P(E

) − P(∪

j<i

)

Substituting this equation into (4.4) yields

P(∪

i=1

) =



P(E

) −



P(∪

j<i

) (4.5)

Because probabilities are always nonnegative, Inequality (4.1) follows directly from

Equation (4.5). Now, ﬁxing i and applying Inequality (4.1) to P(∪

j<i

) yields

P(∪

j<i

) …



j<i

P(E

)

which, by Equation (4.5), gives Inequality (4.2). Similarly, ﬁxing i and applying Inequal-

ity (4.2) to P(∪

j<i

) yields

P(∪

j<i

) Ú



j<i

P(E

) −



k<j<i

P(E

)



j<i

P(E

) −



k<j<i

P(E

)

which, by Equation (4.5), gives Inequality (4.3). The next inclusion–exclusion inequal-

ity is now obtained by ﬁxing i and applying Inequality (4.3) to P(∪

j<i

),and

so on.

2.5 SAMPLE SPACES HAVING EQUALLY LIKELY OUTCOMES

In many experiments, it is natural to assume that all outcomes in the sample space

are equally likely to occur. That is, consider an experiment whose sample space S is a

ﬁnite set, say, S ={1, 2, ..., N}. Then it is often natural to assume that

P({1}) = P({2}) =···=P({N})

which implies, from Axioms 2 and 3 (why?), that

P({i}) =

i = 1, 2, ..., N

34 Chapter 2 Axioms of Probability

From this equation, it follows from Axiom 3 that, for any event E,

P(E) =

number of outcomes in E

number of outcomes in S

In words, if we assume that all outcomes of an experiment are equally likely to occur,

then the probability of any event E equals the proportion of outcomes in the sample

space that are contained in E.

EXAMPLE 5a

If two dice are rolled, what is the probability that the sum of the upturned faces will

equal 7?

Solution. We shall solve this problem under the assumption that all of the 36 possible

outcomes are equally likely. Since there are 6 possible outcomes—namely, (1, 6), (2,

5), (3, 4), (4, 3), (5, 2), and (6, 1)—that result in the sum of the dice being equal to 7,

the desired probability is

. .

EXAMPLE 5b

If 3 balls are “randomly drawn” from a bowl containing 6 white and 5 black balls,

what is the probability that one of the balls is white and the other two black?

Solution. If we regard the order in which the balls are selected as being relevant,

then the sample space consists of 11 · 10 · 9 = 990 outcomes. Furthermore, there are

6 · 5 · 4 = 120 outcomes in which the ﬁrst ball selected is white and the other two

are black; 5 · 6 · 4 = 120 outcomes in which the ﬁrst is black, the second is white,

and the third is black; and 5 · 4 · 6 = 120 in which the ﬁrst two are black and the

third is white. Hence, assuming that “randomly drawn” means that each outcome in

the sample space is equally likely to occur, we see that the desired probability is

120 + 120 + 120

990

This problem could also have been solved by regarding the outcome of the experi-

ment as the unordered set of drawn balls. From this point of view, there are





165 outcomes in the sample space. Now, each set of 3 balls corresponds to 3! out-

comes when the order of selection is noted. As a result, if all outcomes are assumed

equally likely when the order of selection is noted, then it follows that they remain

equally likely when the outcome is taken to be the unordered set of selected balls.

Hence, using the latter representation of the experiment, we see that the desired

probability is











which, of course, agrees with the answer obtained previously.

When the experiment consists of a random selection of k items from a set of n

items, we have the ﬂexibility of either letting the outcome of the experiment be the

ordered selection of the k items or letting it be the unordered set of items selected.

In the former case we would assume that each new selection is equally likely to be

Section 2.5 Sample Spaces Having Equally Likely Outcomes 35

any of the so far unselected items of the set, and in the latter case we would assume

that all





possible subsets of k items are equally likely to be the set selected. For

instance, suppose 5 people are to be randomly selected from a group of 20 individuals

consisting of 10 married couples, and we want to determine P (N), the probability

that the 5 chosen are all unrelated. (That is, no two are married to each other.) If

we regard the sample space as the set of 5 people chosen, then there are





equally

likely outcomes. An outcome that does not contain a married couple can be thought

of as being the result of a six-stage experiment: In the ﬁrst stage, 5 of the 10 couples

to have a member in the group are chosen; in the next 5 stages, 1 of the 2 members

of each of these couples is selected. Thus, there are





possible outcomes in which

the 5 members selected are unrelated, yielding the desired probability of

P(N) =









In contrast, we could let the outcome of the experiment be the ordered selection

of the 5 individuals. In this setting, there are 20 · 19 · 18 · 17 · 16 equally likely

outcomes, of which 20 · 18 · 16 · 14 · 12 outcomes result in a group of 5 unrelated

individuals, yielding the result

P(N) =

20 · 18 · 16 · 14 · 12

20 · 19 · 18 · 17 · 16

We leave it for the reader to verify that the two answers are identical. .

EXAMPLE 5c

A committee of 5 is to be selected from a group of 6 men and 9 women. If the selection

is made randomly, what is the probability that the committee consists of 3 men and 2

women?

Solution. Because each of the





possible committees is equally likely to be selected,

the desired probability is











240

1001

EXAMPLE 5d

An urn contains n balls, one of which is special. If k of these balls are withdrawn one

at a time, with each selection being equally likely to be any of the balls that remain at

the time, what is the probability that the special ball is chosen?

Solution. Since all of the balls are treated in an identical manner, it follows that the

set of k balls selected is equally likely to be any of the





sets of k balls. Therefore,

P{special ball is selected}=





n − 1

k − 1







36 Chapter 2 Axioms of Probability

We could also have obtained this result by letting A

denote the event that the special

ball is the ith ball to be chosen, i = 1, ..., k. Then, since each one of the n balls is

equally likely to be the ith ball chosen, it follows that P(A

) = 1/n. Hence, because

these events are clearly mutually exclusive, we have

P{special ball is selected}=P

⎛

⎝



i=1

⎞

⎠



i=1

P(A

) =

We could also have argued that P(A

) = 1/n, by noting that there are n(n − 1) ···(n −

k + 1) = n!/(n − k)! equally likely outcomes of the experiment, of which (n − 1)(n −

2) ···(n − i + 1)(1)(n − i) ···(n − k + 1) = (n − 1)!/(n − k)! result in the special

ball being the ith one chosen. From this reasoning, it follows that

P(A

) =

(n − 1)!

EXAMPLE 5e

Suppose that n + m balls, of which n are red and m are blue, are arranged in a linear

order in such a way that all (n + m)! possible orderings are equally likely. If we

record the result of this experiment by listing only the colors of the successive balls,

show that all the possible results remain equally likely.

Solution. Consider any one of the (n + m)! possible orderings, and note that any per-

mutation of the red balls among themselves and of the blue balls among themselves

does not change the sequence of colors. As a result, every ordering of colorings cor-

responds to n! m! different orderings of the n + m balls, so every ordering of the

colors has probability

n!m!

(n+m)!

of occurring.

For example, suppose that there are 2 red balls, numbered r

, r

, and 2 blue balls,

numbered b

, b

. Then, of the 4! possible orderings, there will be 2! 2! orderings that

result in any speciﬁed color combination. For instance, the following orderings result

in the successive balls alternating in color, with a red ball ﬁrst:

, b

, r

, b

, r

, b

, r

, b

, r

, b

Therefore, each of the possible orderings of the colors has probability

occurring. .

EXAMPLE 5f

A poker hand consists of 5 cards. If the cards have distinct consecutive values and

are not all of the same suit, we say that the hand is a straight. For instance, a hand

consisting of the ﬁve of spades, six of spades, seven of spades, eight of spades, and

nine of hearts is a straight. What is the probability that one is dealt a straight?

Solution. We start by assuming that all





possible poker hands are equally

likely. To determine the number of outcomes that are straights, let us ﬁrst determine

the number of possible outcomes for which the poker hand consists of an ace, two,

three, four, and ﬁve (the suits being irrelevant). Since the ace can be any 1 of the 4

possible aces, and similarly for the two, three, four, and ﬁve, it follows that there are

outcomes leading to exactly one ace, two, three, four, and ﬁve. Hence, since in 4 of

these outcomes all the cards will be of the same suit (such a hand is called a straight

Section 2.5 Sample Spaces Having Equally Likely Outcomes 37

ﬂush), it follows that there are 4

− 4 hands that make up a straight of the form ace,

two, three, four, and ﬁve. Similarly, there are 4

− 4 hands that make up a straight

of the form ten, jack, queen, king, and ace. Thus, there are 10(4

− 4) hands that are

straights, and it follows that the desired probability is

10(4

− 4)





L .0039 .

EXAMPLE 5g

A 5-card poker hand is said to be a full house if it consists of 3 cards of the same

denomination and 2 other cards of the same denomination (of course, different from

the ﬁrst denomination). Thus, one kind of full house is three of a kind plus a pair.

What is the probability that one is dealt a full house?

Solution. Again, we assume that all





possible hands are equally likely. To

determine the number of possible full houses, we ﬁrst note that there are







different combinations of, say, 2 tens and 3 jacks. Because there are 13 different

choices for the kind of pair and, after a pair has been chosen, there are 12 other

choices for the denomination of the remaining 3 cards, it follows that the probability

of a full house is

13 · 12 ·











L .0014 .

EXAMPLE 5h

In the game of bridge, the entire deck of 52 cards is dealt out to 4 players. What is the

probability that

(a) one of the players receives all 13 spades;

(b) each player receives 1 ace?

Solution. (a) Letting E

be the event that hand i has all 13 spades, then

P(E

) =





, i = 1, 2, 3, 4

Because the events E

, i = 1, 2, 3, 4, are mutually exclusive, the probability that one

of the hands is dealt all 13 spades is

P(∪

i=1

) =



i=1

P(E

) = 4/





L 6.3 * 10

−12

(b) To determine the number of outcomes in which each of the distinct players

receives exactly 1 ace, put aside the aces and note that there are



12, 12, 12, 12



pos-

sible divisions of the other 48 cards when each player is to receive 12. Because there

38 Chapter 2 Axioms of Probability

are 4! ways of dividing the 4 aces so that each player receives 1, we see that the num-

ber of possible outcomes in which each player receives exactly 1 ace is 4!



12, 12, 12, 12



As there are



13,13,13,13



possible hands, the desired probability is thus



12,12,12,12





13,13,13,13



L .1055 .

Some results in probability are quite surprising when initially encountered. Our

next two examples illustrate this phenomenon.

EXAMPLE 5i

If n people are present in a room, what is the probability that no two of them cele-

brate their birthday on the same day of the year? How large need n be so that this

probability is less than

Solution. As each person can celebrate his or her birthday on any one of 365 days,

there is a total of (365)

possible outcomes. (We are ignoring the possibility of some-

one’s having been born on February 29.) Assuming that each outcome is equally

likely, we see that the desired probability is (365)(364)(363)...(365 − n + 1)/(365)

It is a rather surprising fact that when n Ú 23, this probability is less than

. That is, if

there are 23 or more people in a room, then the probability that at least two of them

have the same birthday exceeds

. Many people are initially surprised by this result,

since 23 seems so small in relation to 365, the number of days of the year. However,

every pair of individuals has probability

365

(365)

365

of having the same birthday,

and in a group of 23 people there are





= 253 different pairs of individuals.

Looked at this way, the result no longer seems so surprising.

When there are 50 people in the room, the probability that at least two share the

same birthday is approximately .970, and with 100 persons in the room, the odds are

better than 3,000,000:1. (That is, the probability is greater than

3 * 10

+ 1

that at

least two people have the same birthday.) .

EXAMPLE 5j

A deck of 52 playing cards is shufﬂed, and the cards are turned up one at a time until

the ﬁrst ace appears. Is the next card—that is, the card following the ﬁrst ace—more

likely to be the ace of spades or the two of clubs?

Solution. To determine the probability that the card following the ﬁrst ace is the ace

of spades, we need to calculate how many of the (52)! possible orderings of the cards

have the ace of spades immediately following the ﬁrst ace. To begin, note that each

ordering of the 52 cards can be obtained by ﬁrst ordering the 51 cards different from

the ace of spades and then inserting the ace of spades into that ordering. Furthermore,

for each of the (51)! orderings of the other cards, there is only one place where the

ace of spades can be placed so that it follows the ﬁrst ace. For instance, if the ordering

of the other 51 cards is

4c,6h, Jd,5s, Ac,7d, ..., Kh

Section 2.5 Sample Spaces Having Equally Likely Outcomes 39

then the only insertion of the ace of spades into this ordering that results in its follow-

ing the ﬁrst ace is

4c,6h, Jd,5s, Ac, As,7d, ..., Kh

Therefore, there are (51)! orderings that result in the ace of spades following the ﬁrst

ace, so

P{the ace of spades follows the ﬁrst ace}=

(51)!

(52)!

In fact, by exactly the same argument, it follows that the probability that the two of

clubs (or any other speciﬁed card) follows the ﬁrst ace is also

. In other words, each

of the 52 cards of the deck is equally likely to be the one that follows the ﬁrst ace!

Many people ﬁnd this result rather surprising. Indeed, a common reaction is to

suppose initially that it is more likely that the two of clubs (rather than the ace of

spades) follows the ﬁrst ace, since that ﬁrst ace might itself be the ace of spades. This

reaction is often followed by the realization that the two of clubs might itself appear

before the ﬁrst ace, thus negating its chance of immediately following the ﬁrst ace.

However, as there is one chance in four that the ace of spades will be the ﬁrst ace

(because all 4 aces are equally likely to be ﬁrst) and only one chance in ﬁve that

the two of clubs will appear before the ﬁrst ace (because each of the set of 5 cards

consisting of the two of clubs and the 4 aces is equally likely to be the ﬁrst of this set

to appear), it again appears that the two of clubs is more likely. However, this is not

the case, and a more complete analysis shows that they are equally likely. .

EXAMPLE 5k

A football team consists of 20 offensive and 20 defensive players. The players are to

be paired in groups of 2 for the purpose of determining roommates. If the pairing is

done at random, what is the probability that there are no offensive–defensive room-

mate pairs? What is the probability that there are 2i offensive–defensive roommate

pairs, i = 1, 2, ..., 10?

Solution. There are



2, 2, ...,2



(40)!

(2!)

ways of dividing the 40 players into 20 ordered pairs of two each. [That is, there

are (40)!/2

ways of dividing the players into a ﬁrst pair, a second pair, and so on.]

Hence, there are (40)!/2

(20)! ways of dividing the players into (unordered) pairs of

2 each. Furthermore, since a division will result in no offensive–defensive pairs if the

offensive (and defensive) players are paired among themselves, it follows that there

are [(20)!/2

(10)!]

such divisions. Hence, the probability of no offensive–defensive

roommate pairs, call it P

, is given by



(20)!

(10)!



(40)!

(20)!

[(20)!]

[(10)!]

(40)!

To determine P

, the probability that there are 2i offensive–defensive pairs, we ﬁrst

note that there are





ways of selecting the 2i offensive players and the 2i defen-

sive players who are to be in the offensive–defensive pairs. These 4i players can then

40 Chapter 2 Axioms of Probability

be paired up into (2i)! possible offensive–defensive pairs. (This is so because the ﬁrst

offensive player can be paired with any of the 2i defensive players, the second offen-

sive player with any of the remaining 2i − 1 defensive players, and so on.) As the

remaining 20 − 2i offensive (and defensive) players must be paired among them-

selves, it follows that there are





(2i)!



(20 − 2i)!

10−i

(10 − i)!



divisions which lead to 2i offensive–defensive pairs. Hence,





(2i)!



(20 − 2i)!

10−i

(10 − i)!



(40)!

(20)!

i = 0, 1, ...,10

The P

, i = 0, 1, ..., 10, can now be computed, or they can be approximated by mak-

ing use of a result of Stirling which shows that n! can be approximated by n

n+1/2

−n

√

2π.

For instance, we obtain

L 1.3403 * 10

−6

L .345861

L 7.6068 * 10

−6

Our next three examples illustrate the usefulness of Proposition 4.4. In Example 5l,

the introduction of probability enables us to obtain a quick solution to a counting

problem.

EXAMPLE 5l

A total of 36 members of a club play tennis, 28 play squash, and 18 play badminton.

Furthermore, 22 of the members play both tennis and squash, 12 play both tennis and

badminton, 9 play both squash and badminton, and 4 play all three sports. How many

members of this club play at least one of three sports?

Solution. Let N denote the number of members of the club, and introduce probabil-

ity by assuming that a member of the club is randomly selected. If, for any subset C

of members of the club, we let P(C) denote the probability that the selected member

is contained in C,then

P(C) =

number of members inC

Now, with T being the set of members that plays tennis, S being the set that plays

squash, and B being the set that plays badminton, we have, from Proposition 4.4,

P(T ∪ S ∪ B)

= P(T) + P (S) + P(B) − P(TS) − P (TB) − P(SB) + P (TSB)

36 + 28 + 18 − 22 − 12 − 9 + 4

Hence, we can conclude that 43 members play at least one of the sports. .

Section 2.5 Sample Spaces Having Equally Likely Outcomes 41

The next example in this section not only possesses the virtue of giving rise to a

somewhat surprising answer, but is also of theoretical interest.

EXAMPLE 5m The matching problem

Suppose that each of N men at a party throws his hat into the center of the room.

The hats are ﬁrst mixed up, and then each man randomly selects a hat. What is the

probability that none of the men selects his own hat?

Solution. We ﬁrst calculate the complementary probability of at least one man’s

selecting his own hat. Let us denote by E

, i = 1, 2, ..., N the event that the ith man

selects his own hat. Now, by Proposition 4.4 P





i=1



, the probability that at least

one of the men selects his own hat is given by

⎛

⎝



i=1

⎞

⎠



i=1

P(E

) −



P(E

) + ···

+ (−1)

n+1



···<i

P(E

···E

)

+ ··· + (−1)

N+1

P(E

···E

)

If we regard the outcome of this experiment as a vector of N numbers, where the ith

element is the number of the hat drawn by the ith man, then there are N! possible

outcomes. [The outcome (1, 2, 3, ..., N) means, for example, that each man selects

his own hat.] Furthermore, E

...E

, the event that each of the n men i

, i

, ..., i

selects his own hat, can occur in any of (N − n)(N − n − 1) ···3 · 2 · 1 = (N − n)!

possible ways; for, of the remaining N − n men, the ﬁrst can select any of N − n

hats, the second can then select any of N − n − 1 hats, and so on. Hence, assuming

that all N! possible outcomes are equally likely, we see that

P(E

···E

) =

(N − n)!

Also, as there are





terms in



···<i

P(E

···E

), it follows that



···<i

P(E

···E

) =

N!(N − n)!

(N − n)!n!N!

Thus,

⎛

⎝



i=1

⎞

⎠

= 1 −

− ··· + (−1)

N+1

Hence, the probability that none of the men selects his own hat is

1 − 1 +

−

+ ··· +

(−1)

42 Chapter 2 Axioms of Probability

which is approximately equal to e

−1

L .36788 for N large. In other words, for N large,

the probability that none of the men selects his own hat is approximately .37. (How

many readers would have incorrectly thought that this probability would go to 1 as

→q?) .

For another illustration of the usefulness of Proposition 4.4, consider the following

example.

EXAMPLE 5n

Compute the probability that 10 married couples are seated at random at a round

table, then no wife sits next to her husband.

Solution. If we let E

, i = 1, 2, ..., 10 denote the event that the ith couple sit next

to each other, it follows that the desired probability is 1 − P





i=1



. Now, from

Proposition 4.4,

⎛

⎝



⎞

⎠



P(E

) − ··· + (−1)

n+1



<···<i

P(E

···E

)

+ ··· − P(E

···E

)

To compute P(E

···E

), we ﬁrst note that there are 19! ways of arranging 20

people around a round table. (Why?) The number of arrangements that result in a

speciﬁed set of n men sitting next to their wives can most easily be obtained by ﬁrst

thinking of each of the n married couples as being single entities. If this were the case,

then we would need to arrange 20 − n entities around a round table, and there are

clearly (20 − n − 1)! such arrangements. Finally, since each of the n married couples

can be arranged next to each other in one of two possible ways, it follows that there

are 2

(20 − n − 1)! arrangements that result in a speciﬁed set of n men each sitting

next to their wives. Therefore,

P(E

···E

) =

(19 − n)!

(19)!

Thus, from Proposition 4.4, we obtain that the probability that at least one married

couple sits together, namely,





(18)!

(19)!

−





(17)!

(19)!





(16)!

(19)!

− ···−





(19)!

L .6605

and the desired probability is approximately .3395. .

∗

EXAMPLE 5o Runs

Consider an athletic team that had just ﬁnished its season with a ﬁnal record of n

wins and m losses. By examining the sequence of wins and losses, we are hoping to

determine whether the team had stretches of games in which it was more likely to

win than at other times. One way to gain some insight into this question is to count

the number of runs of wins and then see how likely that result would be when all

Section 2.5 Sample Spaces Having Equally Likely Outcomes 43

(n + m)!/(n! m!) orderings of the n wins and m losses are assumed equally likely. By

a run of wins, we mean a consecutive sequence of wins. For instance, if n = 10, m = 6,

and the sequence of outcomes was WWLLWWWLWLLLWWWW, then there would

be 4 runs of wins—the ﬁrst run being of size 2, the second of size 3, the third of size 1,

and the fourth of size 4.

Suppose now that a team has n wins and m losses. Assuming that all (n + m)!/

(n! m!) =



n + m



orderings are equally likely, let us determine the probability

that there will be exactly r runs of wins. To do so, consider ﬁrst any vector of positive

integers x

, x

, ..., x

with x

+···+x

= n, and let us see how many outcomes result

in r runs of wins in which the ith run is of size x

, i = 1, ..., r . For any such outcome,

if we let y

denote the number of losses before the ﬁrst run of wins, y

the number of

losses between the ﬁrst 2 runs of wins, ..., y

r+1

the number of losses after the last run

of wins, then the y

satisfy

+ y

+ ··· + y

r+1

= my

Ú 0, y

r+1

Ú 0, y

> 0, i = 2, ..., r

and the outcome can be represented schematically as

LL ...L



 

WW ...W



 

L ...L



 

WW ...W



 

··· WW





L ...L



 

r+1

Hence, the number of outcomes that result in r runs of wins—the ith of size x

, i =

1, ...r—is equal to the number of integers y

, ..., y

r+1

that satisfy the foregoing, or,

equivalently, to the number of positive integers

= y

+ 1 y

= y

, i = 2, ..., r, y

r+1

= y

r+1

+ 1

that satisfy

+ y

+ ··· + y

r+1

= m + 2

By Proposition 6.1 in Chapter 1, there are



m + 1



such outcomes. Hence, the

total number of outcomes that result in r runs of wins is



m + 1



, multiplied by

the number of positive integral solutions of x

+ ··· + x

= n. Thus, again from

Proposition 6.1, there are



m + 1



n − 1

r − 1



outcomes resulting in r runs of wins.

As there are



n + m



equally likely outcomes, it follows that

P({r runs of wins}) =



m + 1



n − 1

r − 1





m + n



r Ú 1

For example, if n = 8andm = 6, then the probability of 7 runs is













= 1/429 if all





outcomes are equally likely. Hence, if the outcome

was WLWLWLWLWWLWLW, then we might suspect that the team’s probability

of winning was changing over time. (In particular, the probability that the team wins

44 Chapter 2 Axioms of Probability

seems to be quite high when it lost its last game and quite low when it won its last

game.) On the other extreme, if the outcome were WWWWWWWWLLLLLL,then

there would have been only 1 run, and as P({1run}) =













= 1/429,

it would thus again seem unlikely that the team’s probability of winning remained

unchanged over its 14 games. .

∗

2.6 PROBABILITY AS A CONTINUOUS SET FUNCTION

A sequence of events {E

, n Ú 1} is said to be an increasing sequence if

( E

( ··· ( E

( E

n+1

( ···

whereas it is said to be a decreasing sequence if

) E

) ··· ) E

) E

n+1

) ···

If {E

, n Ú 1}is an increasing sequence of events, then we deﬁne a new event, denoted

by lim

n→q

,by

lim

n→q



i=1

Similarly, if {E

, n Ú 1} is a decreasing sequence of events, we deﬁne lim

lim

n→q



i=1

We now prove the following Proposition 1:

Proposition 6.1.

If {E

, n Ú 1} is either an increasing or a decreasing sequence of events, then

lim

n→q

P(E

) = P( lim

n→q

)

Proof. Suppose, ﬁrst, that {E

, n Ú 1} is an increasing sequence, and deﬁne the

events F

, n Ú 1, by

= E

⎛

⎝

n−1



⎞

⎠

= E

n−1

n > 1

where we have used the fact that

n−1



= E

n−1

, since the events are increasing.

In words, F

consists of those outcomes in E

which are not in any of the earlier

, i < n. It is easy to verify that the F

are mutually exclusive events such that



i=1



i=1

and



i=1



i=1

for all n Ú 1

Section 2.6 Probability as a Continuous Set Function 45

Thus,

⎛

⎝



⎞

⎠

= P

⎛

⎝



⎞

⎠



P(F

)(by Axiom 3)

= lim

n→q



P(F

)

= lim

n→q

⎛

⎝



⎞

⎠

= lim

n→q

⎛

⎝



⎞

⎠

= lim

n→q

P(E

)

which proves the result when {E

, n Ú 1} is increasing.

If {E

, n Ú 1} is a decreasing sequence, then {E

, n Ú 1} is an increasing sequence;

hence, from the preceding equations,

⎛

⎝



⎞

⎠

= lim

n→q

P(E

)

However, because









, it follows that

⎛

⎜

⎝

⎛

⎝



⎞

⎠

⎞

⎟

⎠

= lim

n→q

P(E

)

or, equivalently,

1 − P

⎛

⎝



⎞

⎠

= lim

n→q

[1 − P (E

)] = 1 − lim

n→q

P(E

)

⎛

⎝



⎞

⎠

= lim

n→q

P(E

)

which proves the result.

46 Chapter 2 Axioms of Probability

EXAMPLE 6a Probability and a paradox

Suppose that we possess an inﬁnitely large urn and an inﬁnite collection of balls

labeled ball number 1, number 2, number 3, and so on. Consider an experiment per-

formed as follows: At 1 minute to 12

P.M., balls numbered 1 through 10 are placed

in the urn and ball number 10 is withdrawn. (Assume that the withdrawal takes

no time.) At

minute to 12 P.M., balls numbered 11 through 20 are placed in the

urn and ball number 20 is withdrawn. At

minute to 12 P.M., balls numbered 21

through 30 are placed in the urn and ball number 30 is withdrawn. At

minute

to 12

P.M., and so on. The question of interest is, How many balls are in the urn at

P.M.?

The answer to this question is clearly that there is an inﬁnite number of

balls in the urn at 12

P.M., since any ball whose number is not of the form 10n,

n Ú 1, will have been placed in the urn and will not have been withdrawn before

P.M. Hence, the problem is solved when the experiment is performed as

described.

However, let us now change the experiment and suppose that at 1 minute to 12

P.M.

balls numbered 1 through 10 are placed in the urn and ball number 1 is withdrawn; at

minute to 12 P.M., balls numbered 11 through 20 are placed in the urn and ball num-

ber 2 is withdrawn; at

minute to 12 P.M., balls numbered 21 through 30 are placed

in the urn and ball number 3 is withdrawn; at

minute to 12 P.M., balls numbered 31

through 40 are placed in the urn and ball number 4 is withdrawn, and so on. For this

new experiment, how many balls are in the urn at 12

P.M.?

Surprisingly enough, the answer now is that the urn is empty at 12

P.M. For, consider

any ball—say, ball number n. At some time prior to 12

P.M. [in particular, at





n−1

minutes to 12 P.M.], this ball would have been withdrawn from the urn. Hence, for

each n, ball number n is not in the urn at 12

P.M.; therefore, the urn must be empty at

that time.

We see then, from the preceding discussion that the manner in which the balls are

withdrawn makes a difference. For, in the ﬁrst case only balls numbered 10n, n Ú 1,

are ever withdrawn, whereas in the second case all of the balls are eventually with-

drawn. Let us now suppose that whenever a ball is to be withdrawn, that ball is

randomly selected from among those present. That is, suppose that at 1 minute to

P.M. balls numbered 1 through 10 are placed in the urn and a ball is randomly

selected and withdrawn, and so on. In this case, how many balls are in the urn at

P.M.?

Solution. We shall show that, with probability 1, the urn is empty at 12

P.M. Let us

ﬁrst consider ball number 1. Deﬁne E

to be the event that ball number 1 is still in

the urn after the ﬁrst n withdrawals have been made. Clearly,

P(E

) =

9 · 18 · 27 ···(9n)

10 · 19 · 28 ···(9n + 1)

[To understand this equation, just note that if ball number 1 is still to be in the

urn after the ﬁrst n withdrawals, the ﬁrst ball withdrawn can be any one of 9, the

second any one of 18 (there are 19 balls in the urn at the time of the second with-

drawal, one of which must be ball number 1), and so on. The denominator is similarly

obtained.]

Section 2.6 Probability as a Continuous Set Function 47

Now, the event that ball number 1 is in the urn at 12 P.M. is just the event



n=1

Because the events E

, n Ú 1, are decreasing events, it follows from Proposition 6.1 that

P{ball number 1 is in the urn at 12

P.M.}

= P

⎛

⎝



n=1

⎞

⎠

= lim

n→q

P(E

)



n=1



9n + 1



We now show that



n=1

9n + 1

= 0

Since



n=1



9n + 1



⎡

⎣



n=1



9n + 1



⎤

⎦

−1

this is equivalent to showing that



n=1



1 +



= q

Now, for all m Ú 1,



n=1



1 +





n=1



1 +





1 +



1 +



1 +



···



1 +



+ ··· +



i=1

Hence, letting m

→q and using the fact that



i=1

1/i = q yields



n=1



1 +



= q

Thus, letting F

denote the event that ball number i is in the urn at 12 P.M., we have

shown that P(F

) = 0. Similarly, we can show that P(F

) = 0 for all i.

48 Chapter 2 Axioms of Probability

(For instance, the same reasoning shows that P(F

) =

n=2

[9n/(9n + 1)]fori =

11, 12, ..., 20.) Therefore, the probability that the urn is not empty at 12

P.M., P







satisﬁes

⎛

⎝



⎞

⎠

…



P(F

) = 0

by Boole’s inequality. (See Self-Test Exercise 14.)

Thus, with probability 1, the urn will be empty at 12

P.M. .

2.7 PROBABILITY AS A MEASURE OF BELIEF

Thus far we have interpreted the probability of an event of a given experiment as

being a measure of how frequently the event will occur when the experiment is con-

tinually repeated. However, there are also other uses of the term probability.For

instance, we have all heard such statements as “It is 90 percent probable that Shake-

speare actually wrote Hamlet” or “The probability that Oswald acted alone in assas-

sinating Kennedy is .8.” How are we to interpret these statements?

The most simple and natural interpretation is that the probabilities referred to

are measures of the individual’s degree of belief in the statements that he or she

is making. In other words, the individual making the foregoing statements is quite

certain that Oswald acted alone and is even more certain that Shakespeare wrote

Hamlet. This interpretation of probability as being a measure of the degree of one’s

belief is often referred to as the personal or subjective view of probability.

It seems logical to suppose that a “measure of the degree of one’s belief” should

satisfy all of the axioms of probability. For example, if we are 70 percent certain that

Shakespeare wrote Julius Caesar and 10 percent certain that it was actually Mar-

lowe, then it is logical to suppose that we are 80 percent certain that it was either

Shakespeare or Marlowe. Hence, whether we interpret probability as a measure of

belief or as a long-run frequency of occurrence, its mathematical properties remain

unchanged.

EXAMPLE 7a

Suppose that, in a 7-horse race, you feel that each of the ﬁrst 2 horses has a 20 percent

chance of winning, horses 3 and 4 each have a 15 percent chance, and the remaining

3 horses have a 10 percent chance each. Would it be better for you to wager at even

money that the winner will be one of the ﬁrst three horses or to wager, again at even

money, that the winner will be one of the horses 1, 5, 6, and 7?

Solution. On the basis of your personal probabilities concerning the outcome of

the race, your probability of winning the ﬁrst bet is .2 + .2 + .15 = .55, whereas

it is .2 + .1 + .1 + .1 = .5 for the second bet. Hence, the ﬁrst wager is more

attractive. .

Note that, in supposing that a person’s subjective probabilities are always consis-

tent with the axioms of probability, we are dealing with an idealized rather than an

Summary 49

actual person. For instance, if we were to ask someone what he thought the chances

were of

(a) rain today,

(b) rain tomorrow,

(d) rain either today or tomorrow,

it is quite possible that, after some deliberation, he might give 30 percent, 40 percent,

20 percent, and 60 percent as answers. Unfortunately, such answers (or such subjec-

tive probabilities) are not consistent with the axioms of probability. (Why not?) We

would of course hope that, after this was pointed out to the respondent, she would

change his answers. (One possibility we could accept is 30 percent, 40 percent, 10

percent, and 60 percent.)

SUMMARY

Let S denote the set of all possible outcomes of an experiment. S is called the sample

space of the experiment. An event is a subset of S.IfA

, i = 1, ..., n, are events, then



i=1

, called the union of these events, consists of all outcomes that are in at least

one of the events A

, i = 1, ..., n. Similarly,



i=1

, sometimes written as A

···A

,is

called the intersection of the events A

and consists of all outcomes that are in all of

the events A

, i = 1, ..., n.

For any event A, we deﬁne A

to consist of all outcomes in the sample space that

are not in A. We call A

the complement of the event A. The event S

, which is empty

of outcomes, is designated by Ø and is called the null set. If AB = Ø, then we say that

A and B are mutually exclusive.

For each event A of the sample space S, we suppose that a number P(A), called

the probability of A, is deﬁned and is such that

(i) 0 … P(A) … 1

(ii) P(S) = 1

(iii) For mutually exclusive events A

, i Ú 1,

⎛

⎝



i=1

⎞

⎠



i=1

P(A

)

P(A) represents the probability that the outcome of the experiment is in A.

It can be shown that

P(A

) = 1 − P(A)

A useful result is that

P(A ∪ B) = P(A) + P(B) − P(AB)

50 Chapter 2 Axioms of Probability

which can be generalized to give

⎛

⎝



i=1

⎞

⎠



i=1

P(A

) −



i<j

P(A

) +



i<j<k

P(A

)

+ ··· + (−1)

n+1

P(A

···A

)

If S is ﬁnite and each one point set is assumed to have equal probability, then

P(A) =

|A|

|S|

where |E| denotes the number of outcomes in the event E.

P(A) can be interpreted either as a long-run relative frequency or as a measure of

one’s degree of belief.

PROBLEMS

1. A box contains 3 marbles: 1 red, 1 green, and 1

blue. Consider an experiment that consists of tak-

ing 1 marble from the box and then replacing it

in the box and drawing a second marble from the

box. Describe the sample space. Repeat when the

second marble is drawn without replacing the ﬁrst

marble.

2. In an experiment, die is rolled continually until a

6 appears, at which point the experiment stops.

What is the sample space of this experiment? Let

denote the event that n rolls are necessary to

complete the experiment. What points of the sam-

ple space are contained in E

? What is







3. Two dice are thrown. Let E be the event that the

sum of the dice is odd, let F be the event that

at least one of the dice lands on 1, and let G be

the event that the sum is 5. Describe the events

EF, E ∪ F, FG, EF

,andEFG.

4. A, B,andC take turns ﬂipping a coin. The ﬁrst one

to get a head wins. The sample space of this exper-

iment can be deﬁned by

S =

1, 01, 001, 0001, ...,

0000 ···

(a) Interpret the sample space.

(b) Deﬁne the following events in terms of S:

(i) A wins = A.

(ii) B wins = B.

(iii) (A ∪ B)

Assume that A ﬂips ﬁrst, then B,thenC,

then A, and so on.

5. A system is comprised of 5 components, each of

which is either working or failed. Consider an

experiment that consists of observing the status of

each component, and let the outcome of the exper-

iment be given by the vector (x

, x

where x

is equal to 1 if component i is working

and is equal to 0 if component i is failed.

(a) How many outcomes are in the sample space

of this experiment?

(b) Suppose that the system will work if compo-

nents 1 and 2 are both working, or if compo-

nents 3 and 4 are both working, or if compo-

nents 1, 3, and 5 are all working. Let W be the

event that the system will work. Specify all the

outcomes in W.

are both failed. How many outcomes are con-

tained in the event A?

(d) Write out all the outcomes in the event AW.

6. A hospital administrator codes incoming patients

suffering gunshot wounds according to whether

they have insurance (coding 1 if they do and 0

if they do not) and according to their condition,

which is rated as good (g), fair (f), or serious (s).

Consider an experiment that consists of the coding

of such a patient.

(a) Give the sample space of this experiment.

(b) Let A be the event that the patient is in serious

condition. Specify the outcomes in A.

sured. Specify the outcomes in B.

(d) Give all the outcomes in the event B

∪ A.

7. Consider an experiment that consists of determin-

ing the type of job—either blue-collar or white-

collar—and the political afﬁliation—Republican,

Democratic, or Independent—of the 15 members

of an adult soccer team. How many outcomes are

(a) in the sample space?

(b) in the event that at least one of the team mem-

bers is a blue-collar worker?

Problems 51

considers himself or herself an Independent?

8. Suppose that A and B are mutually exclusive

events for which P(A) = .3andP(B) = .5. What is

the probability that

(a) either A or B occurs?

(b) A occurs but B does not?

9. A retail establishment accepts either the American

Express or the VISA credit card. A total of 24 per-

cent of its customers carry an American Express

card, 61 percent carry a VISA card, and 11 per-

cent carry both cards. What percentage of its cus-

tomers carry a credit card that the establishment

will accept?

10. Sixty percent of the students at a certain school

wear neither a ring nor a necklace. Twenty per-

cent wear a ring and 30 percent wear a necklace.

If one of the students is chosen randomly, what is

the probability that this student is wearing

(a) aringoranecklace?

(b) aringandanecklace?

11. A total of 28 percent of American males smoke

cigarettes, 7 percent smoke cigars, and 5 percent

smoke both cigars and cigarettes.

(a) What percentage of males smokes neither

cigars nor cigarettes?

(b) What percentage smokes cigars but not

cigarettes?

12. An elementary school is offering 3 language

classes: one in Spanish, one in French, and one in

German. The classes are open to any of the 100

students in the school. There are 28 students in the

Spanish class, 26 in the French class, and 16 in the

German class. There are 12 students that are in

both Spanish and French, 4 that are in both Span-

ish and German, and 6 that are in both French and

German. In addition, there are 2 students taking

all 3 classes.

(a) If a student is chosen randomly, what is the

probability that he or she is not in any of the

language classes?

(b) If a student is chosen randomly, what is the

probability that he or she is taking exactly one

language class?

probability that at least 1 is taking a language

class?

13. A certain town with a population of 100,000 has

3 newspapers: I, II, and III. The proportions of

townspeople who read these papers are as follows:

I: 10 percent I and II: 8 percent I and II and

III: 1 percent

II: 30 percent I and III: 2 percent

III: 5 percent II and III: 4 percent

(The list tells us, for instance, that 8000 people

read newspapers I and II.)

(a) Find the number of people who read only one

newspaper.

(b)

How many people read at least two

newspapers?

evening paper, how many people read at least

one morning paper plus an evening paper?

(d) How many people do not read any

newspapers?

(e) How many people read only one morning

paper and one evening paper?

14. The following data were given in a study of a group

of 1000 subscribers to a certain magazine: In ref-

erence to job, marital status, and education, there

were 312 professionals, 470 married persons, 525

college graduates, 42 professional college gradu-

ates, 147 married college graduates, 86 married

professionals, and 25 married professional college

graduates. Show that the numbers reported in the

study must be incorrect.

Hint:LetM, W,andG denote, respectively, the

set of professionals, married persons, and college

graduates. Assume that one of the 1000 persons

is chosen at random, and use Proposition 4.4 to

show that if the given numbers are correct, then

P(M ∪ W ∪ G)>1.

15. If it is assumed that all





poker hands are

equally likely, what is the probability of being dealt

(a) a ﬂush? (A hand is said to be a ﬂush if all 5

cards are of the same suit.)

(b) one pair? (This occurs when the cards have

denominations a, a, b, c, d,wherea, b, c,and

d are all distinct.)

denominations a, a, b, b, c,wherea, b,andc

are all distinct.)

(d) three of a kind? (This occurs when the cards

have denominations a, a

, a, b,

c,wherea, b,

and c are all distinct.)

(e) four of a kind? (This occurs when the cards

have denominations a, a, a, a, b.)

16. Poker dice is played by simultaneously rolling 5

dice. Show that

(a) P{no two alike}=.0926;

(b) P{one pair}=.4630;

(d) P{three alike}=.1543;

(e) P{full house}=.0386;

(f) P{four alike}=.0193;

(g) P{ﬁve alike}=.0008.

17. If 8 rooks (castles) are randomly placed on a

chessboard, compute the probability that none of

the rooks can capture any of the others. That is,

52 Chapter 2 Axioms of Probability

compute the probability that no row or ﬁle con-

tains more than one rook.

18. Two cards are randomly selected from an ordinary

playing deck. What is the probability that they

form a blackjack? That is, what is the probability

that one of the cards is an ace and the other one is

either a ten, a jack, a queen, or a king?

19. Two symmetric dice have both had two of their

sides painted red, two painted black, one painted

yellow, and the other painted white. When this

pair of dice is rolled, what is the probability that

both dice land with the same color face up?

20. Suppose that you are playing blackjack against a

dealer. In a freshly shufﬂed deck, what is the prob-

ability that neither you nor the dealer is dealt a

blackjack?

21. A small community organization consists of 20

families, of which 4 have one child, 8 have two chil-

dren, 5 have three children, 2 have four children,

and 1 has ﬁve children.

(a) If one of these families is chosen at random,

what is the probability it has i children, i =

1, 2, 3, 4, 5?

(b) If one of the children is randomly chosen,

what is the probability that child comes from

a family having i children, i = 1, 2, 3, 4, 5?

22. Consider the following technique for shufﬂing a

deck of n cards: For any initial ordering of the

cards, go through the deck one card at a time and

at each card, ﬂip a fair coin. If the coin comes

up heads, then leave the card where it is; if the

coin comes up tails, then move that card to the

end of the deck. After the coin has been ﬂipped n

times, say that one round has been completed. For

instance, if n = 4 and the initial ordering is 1, 2, 3,

4, then if the successive ﬂips result in the outcome

h, t, t, h, then the ordering at the end of the round

is 1, 4, 2, 3. Assuming that all possible outcomes of

the sequence of n coin ﬂips are equally likely, what

is the probability that the ordering after one round

is the same as the initial ordering?

23. A pair of fair dice is rolled. What is the probabil-

ity that the second die lands on a higher value than

does the ﬁrst?

24. If two dice are rolled, what is the probability that

the sum of the upturned faces equals i?Finditfor

i = 2, 3, ..., 11, 12.

25. A pair of dice is rolled until a sum of either 5 or 7

appears. Find the probability that a 5 occurs ﬁrst.

Hint:LetE

denote the event that a 5 occurs on

the nth roll and no 5 or 7 occurs on the ﬁrst n − 1

rolls. Compute P(E

) and argue that



n=1

P(E

) is

the desired probability.

26. The game of craps is played as follows: A player

rolls two dice. If the sum of the dice is either a 2,

3, or 12, the player loses; if the sum is either a 7

or an 11, the player wins. If the outcome is any-

thing else, the player continues to roll the dice until

she rolls either the initial outcome or a 7. If the 7

comes ﬁrst, the player loses, whereas if the initial

outcome reoccurs before the 7 appears, the player

wins. Compute the probability of a player winning

at craps.

Hint:LetE

denote the event that the initial out-

come is i and the player wins. The desired prob-

ability is



i=2

P(E

). To compute P(E

), deﬁne the

events E

i,n

to be the event that the initial sum is

i and the player wins on the nth roll. Argue that

P(E

) =



n=1

P(E

i,n

27. An urn contains 3 red and 7 black balls. Players A

and B withdraw balls from the urn consecutively

until a red ball is selected. Find the probability that

A selects the red ball. (A draws the ﬁrst ball, then

B, and so on. There is no replacement of the balls

drawn.)

28. An urn contains 5 red, 6 blue, and 8 green balls.

If a set of 3 balls is randomly selected, what is the

probability that each of the balls will be (a) of the

same color? (b) of different colors? Repeat under

the assumption that whenever a ball is selected, its

color is noted and it is then replaced in the urn

before the next selection. This is known as sam-

pling with replacement.

29. An urn contains n white and m black balls, where

n and m are positive numbers.

(a) If two balls are randomly withdrawn, what is

the probability that they are the same color?

(b) If a ball is randomly withdrawn and then

replaced before the second one is drawn, what

is the probability that the withdrawn balls are

the same color?

larger than the one in part (a).

30. The chess clubs of two schools consist of, respec-

tively, 8 and 9 players. Four members from each

club are randomly chosen to participate in a con-

test between the two schools. The chosen play-

ers from one team are then randomly paired with

those from the other team, and each pairing plays

a game of chess. Suppose that Rebecca and her sis-

ter Elise are on the chess clubs at different schools.

What is the probability that

(a) Rebecca and Elise will be paired?

(b) Rebecca and Elise will be chosen to represent

their schools but will not play each other?

represent her school?

Problems 53

31. A 3-person basketball team consists of a guard, a

forward, and a center.

(a) If a person is chosen at random from each of

three different such teams, what is the proba-

bility of selecting a complete team?

(b) What is the probability that all 3 players

selected play the same position?

32. A group of individuals containing b boys and g

girls is lined up in random order; that is, each of

the (b + g)! permutations is assumed to be equally

likely. What is the probability that the person in

the ith position, 1 … i … b + g,isagirl?

33. A forest contains 20 elk, of which 5 are captured,

tagged, and then released. A certain time later, 4

of the 20 elk are captured. What is the probability

that 2 of these 4 have been tagged? What assump-

tions are you making?

34. The second Earl of Yarborough is reported to

have bet at odds of 1000 to 1 that a bridge hand

of 13 cards would contain at least one card that is

ten or higher. (By ten or higher we mean that a

card is either a ten, a jack, a queen, a king, or an

ace.) Nowadays, we call a hand that has no cards

higher than 9 a Yarborough. What is the proba-

bility that a randomly selected bridge hand is a

Yarborough?

35. Seven balls are randomly withdrawn from an urn

that contains 12 red, 16 blue, and 18 green balls.

Find the probability that

(a) 3 red, 2 blue, and 2 green balls are withdrawn;

(b) at least 2 red balls are withdrawn;

(d) either exactly 3 red balls or exactly 3 blue balls

are withdrawn.

36. Two cards are chosen at random from a deck of 52

playing cards. What is the probability that they

(a) are both aces?

(b) have the same value?

37. An instructor gives her class a set of 10 problems

with the information that the ﬁnal exam will con-

sist of a random selection of 5 of them. If a student

has ﬁgured out how to do 7 of the problems, what is

the probability that he or she will answer correctly

(a) all 5 problems?

(b) at least 4 of the problems?

38. There are n socks, 3 of which are red, in a drawer.

What is the value of n if, when 2 of the socks

are chosen randomly, the probability that they are

both red is

39. There are 5 hotels in a certain town. If 3 people

check into hotels in a day, what is the probability

that they each check into a different hotel? What

assumptions are you making?

40. A town contains 4 people who repair televisions.

If 4 sets break down, what is the probability that

exactly i of the repairers are called? Solve the

problem for i = 1, 2, 3, 4. What assumptions are

you making?

41. If a die is rolled 4 times, what is the probability that

6 comes up at least once?

42. Two dice are thrown n times in succession. Com-

pute the probability that double 6 appears at least

once. How large need n be to make this probability

at least

43. (a) If N people, including A and B, are randomly

arranged in a line, what is the probability that

A and B are next to each other?

(b) What would the probability be if the people

were randomly arranged in a circle?

44. Five people, designated as A, B, C, D, E,are

arranged in linear order. Assuming that each pos-

sible order is equally likely, what is the probabil-

ity that

(a) there is exactly one person between A and B?

(b) there are exactly two people between A

and B?

45. A woman has n keys, of which one will open

her door.

(a) If she tries the keys at random, discarding

those that do not work, what is the probability

that she will open the door on her kth try?

(b) What if she does not discard previously tried

keys?

46. How many people have to be in a room in order

that the probability that at least two of them cele-

brate their birthday in the same month is at least

? Assume that all possible monthly outcomes are

equally likely.

47. If there are 12 strangers in a room, what is the

probability that no two of them celebrate their

birthday in the same month?

48. Given 20 people, what is the probability that,

among the 12 months in the year, there are 4

months containing exactly 2 birthdays and 4 con-

taining exactly 3 birthdays?

49. A group of 6 men and 6 women is randomly

divided into 2 groups of size 6 each. What is the

probability that both groups will have the same

number of men?

50. In a hand of bridge, ﬁnd the probability that you

have 5 spades and your partner has the remain-

ing 8.

51. Suppose that n balls are randomly distributed into

N compartments. Find the probability that m balls

will fall into the ﬁrst compartment. Assume that all

arrangements are equally likely.

54 Chapter 2 Axioms of Probability

52. A closet contains 10 pairs of shoes. If 8 shoes

are randomly selected, what is the probability that

there will be

(a) no complete pair?

(b) exactly 1 complete pair?

53. If 4 married couples are arranged in a row, ﬁnd the

probability that no husband sits next to his wife.

54. Compute the probability that a bridge hand is void

in at least one suit. Note that the answer is not











(Why not?)

Hint: Use Proposition 4.4.

55. Compute the probability that a hand of 13 cards

contains

(a) the ace and king of at least one suit;

(b) all 4 of at least 1 of the 13 denominations.

56. Two players play the following game: Player A

chooses one of the three spinners pictured in

Figure 2.6, and then player B chooses one of the

remaining two spinners. Both players then spin

their spinner, and the one that lands on the higher

number is declared the winner. Assuming that

each spinner is equally likely to land in any of its

3 regions, would you rather be player A or player

B? Explain your answer!

FIGURE 2.6: Spinners

THEORETICAL EXERCISES

Prove the following relations:

1. EF ( E ( E ∪ F.

2. If E ( F,thenF

( E

3. F = FE ∪ FE

and E ∪ F = E ∪ E







F =



F and







∪ F =



∪ F).

Theoretical Exercises 55

5. For any sequence of events E

, E

, ..., deﬁne a

new sequence F

, F

, ... of disjoint events (that is,

events such that F

= Ø whenever i Z j)such

that for all n Ú 1,



6. Let E, F,andG be three events. Find expressions

for the events so that, of E, F,andG,

(a) only E occurs;

(b) both E and G, but not F, occur;

(d) at least two of the events occur;

(e) all three events occur;

(f) none of the events occurs;

(g) at most one of the events occurs;

(h) at most two of the events occur;

(i) exactly two of the events occur;

(j) at most three of the events occur.

7. Find the simplest expression for the following

events:

(a) (E ∪ F)(E ∪ F

);

(b) (E ∪ F)(E

∪ F)(E ∪ F

);

8. Let S be a given set. If, for some k > 0,

, S

, ..., S

are mutually exclusive nonempty

subsets of S such that



i=1

= S,thenwe

call the set {S

, S

, ..., S

} a partition of S.Let

denote the number of different partitions of

{1, 2, ..., n}. Thus, T

= 1 (the only partition being

={1})andT

= 2 (the two partitions being

{{1, 2, }}, {{1}, {2}}).

(a) Show, by computing all partitions, that T

5, T

= 15.

(b) Show that

n+1

= 1 +



k=1





and use this equation to compute T

Hint: One way of choosing a partition of n + 1

items is to call one of the items special.Thenwe

obtain different partitions by ﬁrst choosing k, k =

0, 1, ..., n, then a subset of size n − k of the non-

special items, and then any of the T

partitions of

the remaining k nonspecial items. By adding the

special item to the subset of size n − k, we obtain

a partition of all n + 1 items.

9. Suppose that an experiment is performed n times.

For any event E of the sample space, let n(E)

denote the number of times that event E occurs

and deﬁne f (E) = n(E)/n. Show that f(·) satisﬁes

Axioms 1, 2, and 3.

10. Prove that P(E ∪ F ∪ G) = P(E) + P(F) +

P(G) − P(E

FG) − P(EF

G) − P(EFG

) −

2P(EFG).

11. If P(E) = .9andP(F) = .8, show that P(EF) Ú .7.

In general, prove Bonferroni’s inequality, namely,

P(EF) Ú P(E) + P(F) − 1

12. Show that the probability that exactly one of the

events E or F occurs equals P(E) + P(F) −

2P(EF).

13. Prove that P(EF

) = P(E) − P(EF).

14. Prove Proposition 4.4 by mathematical induction.

15. An urn contains M white and N black balls. If a

random sample of size r is chosen, what is the prob-

ability that it contains exactly k white balls?

16. Use induction to generalize Bonferroni’s inequal-

ity to n events. That is, show that

P(E

···E

) Ú P(E

) +···+P(E

) − (n − 1)

17. Consider the matching problem, Example 5m, and

deﬁne A

to be the number of ways in which the

N men can select their hats so that no man selects

his own. Argue that

= (N − 1)(A

N−1

+ A

N−2

)

This formula, along with the boundary conditions

= 0, A

= 1, can then be solved for A

,and

the desired probability of no matches would be

/N!.

Hint: After the ﬁrst man selects a hat that is not his

own, there remain N − 1 men to select among a set

of N − 1 hats that does not contain the hat of one

of these men. Thus, there is one extra man and one

extra hat. Argue that we can get no matches either

with the extra man selecting the extra hat or with

the extra man not selecting the extra hat.

18. Let f

denote the number of ways of tossing a coin

n times such that successive heads never appear.

Argue that

= f

n−1

+ f

n−2

n Ú 2, where f

K 1, f

K 2

Hint: How many outcomes are there that start

with a head, and how many start with a tail? If

denotes the probability that successive heads

never appear when a coin is tossed n times, ﬁnd

(in terms of f

) when all possible outcomes of

the n tosses are assumed equally likely. Compute

19. An urn contains n red and m blue balls. They are

withdrawn one at a time until a total of r, r … n,

56 Chapter 2 Axioms of Probability

red balls have been withdrawn. Find the probabil-

ity that a total of k balls are withdrawn.

Hint: A total of k balls will be withdrawn if there

are r − 1 red balls in the ﬁrst k − 1 withdrawals

and the kth withdrawal is a red ball.

20. Consider an experiment whose sample space con-

sists of a countably inﬁnite number of points.

Show that not all points can be equally likely.

Can all points have a positive probability of

occurring?

∗

21. Consider Example 5o, which is concerned with the

number of runs of wins obtained when n wins and

m losses are randomly permuted. Now

consider the total number of runs—that is, win

runs plus loss runs—and show that

P{2k runs}=2



m − 1

k − 1



n − 1

k − 1





m + n



P{2k + 1 runs}



m − 1

k − 1



n − 1





m − 1



n − 1

k − 1





m + n



SELF-TEST PROBLEMS AND EXERCISES

1. A cafeteria offers a three-course meal consisting

of an entree, a starch, and a dessert. The possible

choices are given in the following table:

Course Choices

Entree Chicken or roast beef

Starch Pasta or rice or potatoes

Dessert Ice cream or Jello or apple pie or a peach

A person is to choose one course from each cate-

gory.

(a) How many outcomes are in the sample space?

(b) Let A be the event that ice cream is chosen.

How many outcomes are in A?

How many outcomes are in B?

(d) List all the outcomes in the event AB.

(e) Let C be the event that rice is chosen. How

many outcomes are in C?

(f) List all the outcomes in the event ABC.

2. A customer visiting the suit department of a cer-

tain store will purchase a suit with probability .22,

a shirt with probability .30, and a tie with proba-

bility .28. The customer will purchase both a suit

and a shirt with probability .11, both a suit and a

tie with probability .14, and both a shirt and a tie

with probability .10. A customer will purchase all 3

items with probability .06. What is the probability

that a customer purchases

(a) none of these items?

(b) exactly 1 of these items?

3. A deck of cards is dealt out. What is the proba-

bility that the 14th card dealt is an ace? What is

the probability that the ﬁrst ace occurs on the 14th

card?

4. Let A denote the event that the midtown temper-

ature in Los Angeles is 70

◦

F, and let B denote the

event that the midtown temperature in New York

is 70

◦

F. Also, let C denote the event that the max-

imum of the midtown temperatures in New York

and in Los Angeles is 70

◦

F. If P(A) = .3, P(B) =

.4, and P(C) = .2, ﬁnd the probability that the min-

imum of the two midtown temperatures is 70

◦

5. An ordinary deck of 52 cards is shufﬂed. What is

the probability that the top four cards have

(a) different denominations?

(b) different suits?

6. Urn A contains 3 red and 3 black balls, whereas

urn B contains 4 red and 6 black balls. If a ball is

randomly selected from each urn, what is the prob-

ability that the balls will be the same color?

7. In a state lottery, a player must choose 8 of the

numbers from 1 to 40. The lottery commission

then performs an experiment that selects 8 of these

40 numbers. Assuming that the choice of the lot-

tery commission is equally likely to be any of the





combinations, what is the probability that a

player has

(a) all 8 of the numbers selected by the lottery

commission?

(b) 7 of the numbers selected by the lottery com-

mission?

tery commission?

8. From a group of 3 freshmen, 4 sophomores, 4

juniors, and 3 seniors a committee of size 4 is ran-

domly selected. Find the probability that the com-

mittee will consist of

(a) 1 from each class;

(b) 2 sophomores and 2 juniors;

Self -Test Problems and Exercises 57

9. For a ﬁnite set A,letN(A) denote the number of

elements in A.

(a) Show that

N(A ∪ B) = N(A) + N(B) − N(AB)

(b) More generally, show that

⎛

⎝



i=1

⎞

⎠



N(A

) −



i<j

N(A

)

+ ··· + (−1)

n+1

N(A

···A

)

10. Consider an experiment that consists of six horses,

numbered 1 through 6, running a race, and sup-

pose that the sample space consists of the 6! pos-

sible orders in which the horses ﬁnish. Let A be

the event that the number-1 horse is among the

top three ﬁnishers, and let B be the event that the

number-2 horse comes in second. How many out-

comes are in the event A ∪ B?

11. A 5-card hand is dealt from a well-shufﬂed deck of

52 playing cards. What is the probability that the

hand contains at least one card from each of the

four suits?

12. A basketball team consists of 6 frontcourt and

4 backcourt players. If players are divided into

roommates at random, what is the probability that

there will be exactly two roommate pairs made up

of a backcourt and a frontcourt player?

13. Suppose that a person chooses a letter at random

from R E S E R V E and then chooses one at ran-

dom from V E R T I C A L. What is the probability

that the same letter is chosen?

14. Prove Boole’s inequality:

⎛

⎝



i=1

⎞

⎠

…



i=1

P(A

)

15. Show that if P(A

) = 1foralli Ú 1, then





i=1



= 1.

16. Let T

(n) denote the number of partitions of the

set {1, ..., n} into k nonempty subsets, where 1 …

k … n. (See Theoretical Exercise 8 for the deﬁni-

tion of a partition.) Argue that

(n) = kT

(n − 1) + T

k−1

(n − 1)

Hint: In how many partitions is {1} a subset, and in

how many is 1 an element of a subset that contains

other elements?

17. Five balls are randomly chosen, without replace-

ment, from an urn that contains 5 red, 6 white, and

7 blue balls. Find the probability that at least one

ball of each color is chosen.

18. Four red, 8 blue, and 5 green balls are randomly

arranged in a line.

(a) What is the probability that the ﬁrst 5 balls are

blue?

(b) What is the probability that none of the ﬁrst 5

balls are blue?

are differently colored.

(d) What is the probability that all the red balls

are together?

19. Ten cards are randomly chosen from a deck of 52

cards that consists of 13 cards of each of 4 different

suits. Each of the selected cards is put in one of 4

piles, depending on the suit of the card.

(a) What is the probability that the largest pile

has 4 cards, the next largest has 3, the next

largest has 2, and the smallest has 1 card?

(b) What is the probability that two of the piles

have 3 cards, one has 4 cards, and one has no

cards?

20. Balls are randomly removed from an urn initially

containing 20 red and 10 blue balls. What is the

probability that all of the red balls are removed

before all of the blue ones have been removed?

CHAPTER 3

Conditional Probability and Independence

3.1 INTRODUCTION

3.2 CONDITIONAL PROBABILITIES

3.3 BAYES’S FORMULA

3.4 INDEPENDENT EVENTS

3.5 P(·|F ) IS A PROBABILITY

3.1 INTRODUCTION

In this chapter, we introduce one of the most important concepts in probability theory,

that of conditional probability. The importance of this concept is twofold. In the ﬁrst

place, we are often interested in calculating probabilities when some partial informa-

tion concerning the result of an experiment is available; in such a situation, the desired

probabilities are conditional. Second, even when no partial information is available,

conditional probabilities can often be used to compute the desired probabilities more

easily.

3.2 CONDITIONAL PROBABILITIES

Suppose that we toss 2 dice, and suppose that each of the 36 possible outcomes is

equally likely to occur and hence has probability

. Suppose further that we observe

that the ﬁrst die is a 3. Then, given this information, what is the probability that the

sum of the 2 dice equals 8? To calculate this probability, we reason as follows: Given

that the initial die is a 3, there can be at most 6 possible outcomes of our experiment,

namely, (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), and (3, 6). Since each of these outcomes

originally had the same probability of occurring, the outcomes should still have equal

probabilities. That is, given that the ﬁrst die is a 3, the (conditional) probability of

each of the outcomes (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), and (3, 6) is

, whereas the

(conditional) probability of the other 30 points in the sample space is 0. Hence, the

desired probability will be

If we let E and F denote, respectively, the event that the sum of the dice is 8 and the

event that the ﬁrst die is a 3, then the probability just obtained is called the conditional

probability that E occurs given that F has occurred and is denoted by

P(E|F)

A general formula for P(E|F) that is valid for all events E and F is derived in the same

manner: If the event F occurs, then, in order for E to occur, it is necessary that the

actual occurrence be a point both in E and in F; that is, it must be in EF. Now, since

we know that F has occurred, it follows that F becomes our new, or reduced, sample

space; hence, the probability that the event EF occurs will equal the probability of

EF relative to the probability of F. That is, we have the following deﬁnition.

Section 3.2 Conditional Probabilities 59

Deﬁnition

If P(F)>0, then

P(E|F) =

P(EF)

P(F)

(2.1)

EXAMPLE 2a

A student is taking a one-hour-time-limit makeup examination. Suppose the proba-

bility that the student will ﬁnish the exam in less than x hours is x/2, for all 0 … x … 1.

Then, given that the student is still working after .75 hour, what is the conditional

probability that the full hour is used?

Solution. Let L

denote the event that the student ﬁnishes the exam in less than x

hours, 0 … x … 1, and let F be the event that the student uses the full hour. Because

F is the event that the student is not ﬁnished in less than 1 hour,

P(F) = P(L

) = 1 − P(L

) = .5

Now, the event that the student is still working at time .75 is the complement of the

event L

.75

, so the desired probability is obtained from

P(F|L

.75

) =

P(FL

.75

)

P(L

.75

)

P(F)

1 − P(L

.75

)

.625

= .8

If each outcome of a ﬁnite sample space S is equally likely, then, conditional on

the event that the outcome lies in a subset F ( S, all outcomes in F become equally

likely. In such cases, it is often convenient to compute conditional probabilities of

the form P(E|F) by using F as the sample space. Indeed, working with this reduced

sample space often results in an easier and better understood solution. Our next few

examples illustrate this point.

EXAMPLE 2b

A coin is ﬂipped twice. Assuming that all four points in the sample space S ={(h, h),

(h, t), (t, h), (t, t)} are equally likely, what is the conditional probability that both ﬂips

land on heads, given that (a) the ﬁrst ﬂip lands on heads? (b) at least one ﬂip lands

on heads?

Solution. Let B ={(h, h)} be the event that both ﬂips land on heads; let F ={(h, h),

(h, t)} be the event that the ﬁrst ﬂip lands on heads; and let A ={(h, h), (h, t), (t

, h)} be

the event that at least one ﬂip lands on heads. The probability for (a) can be obtained

from

60 Chapter 3 Conditional Probability and Independence

P(B|F) =

P(BF)

P(F)

P({(h, h)})

P({(h, h), (h, t)})

1/4

2/4

= 1/2

For (b), we have

P(B|A) =

P(BA)

P(A)

P({(h, h)})

P({(h, h), (h, t), (t, h)})

1/4

3/4

= 1/3

Thus, the conditional probability that both ﬂips land on heads given that the ﬁrst

one does is 1/2, whereas the conditional probability that both ﬂips land on heads

given that at least one does is only 1/3. Many students initially ﬁnd this latter result

surprising. They reason that, given that at least one ﬂip lands on heads, there are

two possible results: Either they both land on heads or only one does. Their mistake,

however, is in assuming that these two possibilities are equally likely. For, initially,

there are 4 equally likely outcomes. Because the information that at least one ﬂip

lands on heads is equivalent to the information that the outcome is not (t, t), we are

left with the 3 equally likely outcomes (h, h), (h, t), (t, h), only one of which results in

both ﬂips landing on heads. .

EXAMPLE 2c

In the card game bridge, the 52 cards are dealt out equally to 4 players—called East,

West, North, and South. If North and South have a total of 8 spades among them,

what is the probability that East has 3 of the remaining 5 spades?

Solution. Probably the easiest way to compute the desired probability is to work

with the reduced sample space. That is, given that North–South have a total of 8

spades among their 26 cards, there remains a total of 26 cards, exactly 5 of them

being spades, to be distributed among the East–West hands. Since each distribution

is equally likely, it follows that the conditional probability that East will have exactly

3 spades among his or her 13 cards is











L .339

EXAMPLE 2d

A total of n balls are sequentially and randomly chosen, without replacement, from

an urn containing r red and b blue balls (n … r + b). Given that k of the n balls are

blue, what is the conditional probability that the ﬁrst ball chosen is blue?

Section 3.2 Conditional Probabilities 61

Solution. If we imagine that the balls are numbered, with the blue balls having num-

bers 1 through b and the red balls b + 1 through b + r, then the outcome of the

experiment of selecting n balls without replacement is a vector of distinct integers

, ..., x

, where each x

is between 1 and r + b. Moreover, each such vector is equally

likely to be the outcome. So, given that the vector contains k blue balls (that is, it

contains k values between 1 and b), it follows that each of these outcomes is equally

likely. But because the ﬁrst ball chosen is, therefore, equally likely to be any of the n

chosen balls, of which k are blue, it follows that the desired probability is k/n.

If we did not choose to work with the reduced sample space, we could have solved

the problem by letting B be the event that the ﬁrst ball chosen is blue and B

be the

event that a total of k blue balls are chosen. Then

P(B|B

) =

P(BB

)

P(B

)

P(B

|B)P(B)

P(B

)

Now, P(B

|B) is the probability that a random choice of n − 1 balls from an urn

containing r red and b − 1 blue balls results in a total of k − 1 blue balls being

chosen; consequently,

P(B

|B) =



b−1

k−1



n−k





r+b−1

n−1



Using the preceding formula along with

P(B) =

r + b

and the hypergeometric probability

P(B

) =





n−k





r+b



again yields the result that

P(B|B

) =

Multiplying both sides of Equation (2.1) by P(F ), we obtain

P(EF ) = P(F)P(E|F ) (2.2)

In words, Equation (2.2) states that the probability that both E and F occur is equal

to the probability that F occurs multiplied by the conditional probability of E given

that F occurred. Equation (2.2) is often quite useful in computing the probability of

the intersection of events.

EXAMPLE 2e

Celine is undecided as to whether to take a French course or a chemistry course. She

estimates that her probability of receiving an A grade would be

in a French course

and

in a chemistry course. If Celine decides to base her decision on the ﬂip of a fair

coin, what is the probability that she gets an A in chemistry?

62 Chapter 3 Conditional Probability and Independence

Solution. (a) Let the event that Celine takes chemistry and A denote the event that

she receives an A in whatever course she takes, then the desired probability is P(CA),

which is calculated by using Equation (2.2) as follows:

P(CA) = P(C)P(A|C)







EXAMPLE 2f

Suppose that an urn contains 8 red balls and 4 white balls. We draw 2 balls from the

urn without replacement. (a) If we assume that at each draw each ball in the urn is

equally likely to be chosen, what is the probability that both balls drawn are red? (b)

Now suppose that the balls have different weights, with each red ball having weight r

and each white ball having weight w. Suppose that the probability that a given ball in

the urn is the next one selected is its weight divided by the sum of the weights of all

balls currently in the urn. Now what is the probability that both balls are red?

Solution. Let R

and R

denote, respectively, the events that the ﬁrst and second

balls drawn are red. Now, given that the ﬁrst ball selected is red, there are 7 remaining

red balls and 4 white balls, so P(R

) =

.AsP(R

) is clearly

, the desired

probability is

P(R

) = P(R

)P(R

)







Of course, this probability could have been computed by P(R

) =









For part (b), we again let R

be the event that the ith ball chosen is red and use

P(R

) = P(R

)P(R

)

Now, number the red balls, and let B

, i = 1, ..., 8 be the event that the ﬁrst ball

drawn is red ball number i. Then

P(R

) = P(∪

i=1

) =



i=1

P(B

) = 8

8r + 4w

Moreover, given that the ﬁrst ball is red, the urn then contains 7 red and 4 white balls.

Thus, by an argument similar to the preceding one,

P(R

) =

7r + 4w

Hence, the probability that both balls are red is

P(R

) =

8r + 4w

7r + 4w

A generalization of Equation (2.2), which provides an expression for the probabil-

ity of the intersection of an arbitrary number of events, is sometimes referred to as

the multiplication rule.

Section 3.2 Conditional Probabilities 63

The multiplication rule

P(E

···E

) = P(E

)P(E

) ···P(E

···E

n−1

)

To prove the multiplication rule, just apply the deﬁnition of conditional probability

to its right-hand side, giving

P(E

)

P(E

)

P(E

)

P(E

)

P(E

)

···

P(E

···E

)

P(E

···E

n−1

)

= P(E

···E

)

EXAMPLE 2g

In the match problem stated in Example 5m of Chapter 2, it was shown that P

,the

probability that there are no matches when N people randomly select from among

their own N hats, is given by



i=0

(−1)

/i!

What is the probability that exactly k of the N people have matches?

Solution. Let us ﬁx our attention on a particular set of k people and determine the

probability that these k individuals have matches and no one else does. Letting E

denote the event that everyone in this set has a match, and letting G be the event that

none of the other N − k people have a match, we have

P(EG) = P(E )P(G|E )

Now, let F

, i = 1, ..., k, be the event that the ith member of the set has a match.

Then

P(E ) = P(F

···F

)

= P(F

)P(F

) ···P(F

···F

k−1

)

N − 1

N − 2

···

N − k + 1

(N − k)!

Given that everyone in the set of k has a match, the other N − k people will be

randomly choosing among their own N − k hats, so the probability that none of

them has a match is equal to the probability of no matches in a problem having N − k

people choosing among their own N − k hats. Therefore,

P(G|E ) = P

N−k



i=0

(−1)

/i!

showing that the probability that a speciﬁed set of k people have matches and no one

else does is

P(EG) =

(N − k)!

N−k

64 Chapter 3 Conditional Probability and Independence

Because there will be exactly k matches if the preceding is true for any of the





sets

of k individuals, the desired probability is

P(exactly k matches) = P

N−k

/k!

L e

−1

/k! when N is large

We will now employ the multiplication rule to obtain a second approach to solving

Example 5h(b) of Chapter 2.

EXAMPLE 2h

An ordinary deck of 52 playing cards is randomly divided into 4 piles of 13 cards each.

Compute the probability that each pile has exactly 1 ace.

Solution. Deﬁne events E

, i = 1, 2, 3, 4, as follows:

={the ace of spades is in any one of the piles}

={the ace of spades and the ace of hearts are in different piles}

={the aces of spades, hearts, and diamonds are all in different piles}

={all 4 aces are in different piles}

The desired probability is P(E

), and by the multiplication rule,

P(E

) = P(E

)P(E

)

Now,

P(E

) = 1

since E

is the sample space S. Also,

P(E

) =

since the pile containing the ace of spades will receive 12 of the remaining 51 cards,

and

P(E

) =

since the piles containing the aces of spades and hearts will receive 24 of the remain-

ing 50 cards. Finally,

P(E

) =

Therefore, the probability that each pile has exactly 1 ace is

P(E

) =

39 · 26 · 13

51 · 50 · 49

L .105

That is, there is approximately a 10.5 percent chance that each pile will contain an

ace. (Problem 13 gives another way of using the multiplication rule to solve this

problem.)

Remarks. Our deﬁnition of P(E|F) is consistent with the interpretation of prob-

ability as being a long-run relative frequency. To see this, suppose that n repeti-

tions of the experiment are to be performed, where n is large. We claim that if

Section 3.3 Bayes’s Formula 65

we consider only those experiments in which F occurs, then P(E|F) will equal the

long-run proportion of them in which E also occurs. To verify this statement, note

that, since P(F) is the long-run proportion of experiments in which F occurs,

it follows that in the n repetitions of the experiment F will occur approximately

nP(F) times. Similarly, in approximately nP (EF) of these experiments both E and

F will occur. Hence, out of the approximately nP(F) experiments in which

F occurs, the proportion of them in which E also occurs is approximately

equal to

nP(EF)

nP(F)

P(EF)

P(F)

Because this approximation becomes exact as n becomes larger and larger, we have

the appropriate deﬁnition of P(E|F).

3.3 BAYES’S FORMULA

Let E and F be events. We may express E as

E = EF ∪ EF

for, in order for an outcome to be in E, it must either be in both E and F or be in

E but not in F. (See Figure 3.1.) As EF and EF

are clearly mutually exclusive, we

have, by Axiom 3,

P(E ) = P(EF ) + P(EF

)

= P(E|F )P(F ) + P(E|F

)P(F

)

= P(E|F )P(F ) + P(E|F

)[1 − P (F)]

(3.1)

Equation (3.1) states that the probability of the event E is a weighted average of the

conditional probability of E given that F has occurred and the conditional proba-

bility of E given that F has not occurred—each conditional probability being given

as much weight as the event on which it is conditioned has of occurring. This is an

extremely useful formula, because its use often enables us to determine the prob-

ability of an event by ﬁrst “conditioning” upon whether or not some second event

has occurred. That is, there are many instances in which it is difﬁcult to compute the

probability of an event directly, but it is straightforward to compute it once we know

whether or not some second event has occurred. We illustrate this idea with some

examples.

EFEF

FIGURE 3.1: E = EF ∪ EF

. EF = Shaded Area; EF

= Striped Area

66 Chapter 3 Conditional Probability and Independence

EXAMPLE 3a (Part 1)

An insurance company believes that people can be divided into two classes: those

who are accident prone and those who are not. The company’s statistics show that

an accident-prone person will have an accident at some time within a ﬁxed 1-year

period with probability .4, whereas this probability decreases to .2 for a person who is

not accident prone. If we assume that 30 percent of the population is accident prone,

what is the probability that a new policyholder will have an accident within a year of

purchasing a policy?

Solution. We shall obtain the desired probability by ﬁrst conditioning upon whether

or not the policyholder is accident prone. Let A

denote the event that the policy-

holder will have an accident within a year of purchasing the policy, and let A denote

the event that the policyholder is accident prone. Hence, the desired probability is

given by

P(A

) = P(A

|A)P(A) + P(A

)P(A

)

= (.4)(.3) + (.2)(.7) = .26

EXAMPLE 3a (Part 2)

Suppose that a new policyholder has an accident within a year of purchasing a policy.

What is the probability that he or she is accident prone?

Solution. The desired probability is

P(A|A

) =

P(AA

)

P(A

)

P(A)P(A

|A)

P(A

)

(.3)(.4)

.26

EXAMPLE 3b

Consider the following game played with an ordinary deck of 52 playing cards: The

cards are shufﬂed and then turned over one at a time. At any time, the player can

guess that the next card to be turned over will be the ace of spades; if it is, then the

player wins. In addition, the player is said to win if the ace of spades has not yet

appeared when only one card remains and no guess has yet been made. What is a

good strategy? What is a bad strategy?

Solution. Every strategy has probability 1/52 of winning! To show this, we will use

induction to prove the stronger result that, for an n card deck, one of whose cards

is the ace of spades, the probability of winning is 1/n, no matter what strategy is

employed. Since this is clearly true for n = 1, assume it to be true for an n − 1

card deck, and now consider an n card deck. Fix any strategy, and let p denote the

probability that the strategy guesses that the ﬁrst card is the ace of spades. Given that

it does, the player’s probability of winning is 1/n. If, however, the strategy does not

guess that the ﬁrst card is the ace of spades, then the probability that the player wins

is the probability that the ﬁrst card is not the ace of spades, namely, (n − 1)/n, multi-

plied by the conditional probability of winning given that the ﬁrst card is not the ace

of spades. But this latter conditional probability is equal to the probability of winning

Section 3.3 Bayes’s Formula 67

whenusingann − 1 card deck containing a single ace of spades; it is thus, by the

induction hypothesis, 1/(n − 1). Hence, given that the strategy does not guess the

ﬁrst card, the probability of winning is

n − 1

Thus, letting G be the event that the ﬁrst card is guessed, we obtain

P{win}=P{win|G}P(G) + P{win|G

}(1 − P(G)) =

p +

(1 − p)

EXAMPLE 3c

In answering a question on a multiple-choice test, a student either knows the answer

or guesses. Let p be the probability that the student knows the answer and 1 − p

be the probability that the student guesses. Assume that a student who guesses at the

answer will be correct with probability 1/m, where m is the number of multiple-choice

alternatives. What is the conditional probability that a student knew the answer to a

question given that he or she answered it correctly?

Solution. Let C and K denote, respectively, the events that the student answers the

question correctly and the event that he or she actually knows the answer. Now,

P(K|C ) =

P(KC )

P(C )

P(C|K )P(K )

P(C|K )P(K ) + P(C|K

)P(K

)

p + (1/m)(1 − p)

1 + (m − 1)p

For example, if m = 5, p =

, then the probability that the student knew the answer

to a question he or she answered correctly is

EXAMPLE 3d

A laboratory blood test is 95 percent effective in detecting a certain disease when it

is, in fact, present. However, the test also yields a “false positive” result for 1 percent

of the healthy persons tested. (That is, if a healthy person is tested, then, with prob-

ability .01, the test result will imply that he or she has the disease.) If .5 percent of

the population actually has the disease, what is the probability that a person has the

disease given that the test result is positive?

68 Chapter 3 Conditional Probability and Independence

Solution. Let D be the event that the person tested has the disease and E the event

that the test result is positive. Then the desired probability is

P(D|E) =

P(DE)

P(E)

P(E|D)P(D)

P(E|D)P(D) + P(E|D

)P(D

)

(.95)(.005)

(.95)(.005) + (.01)(.995)

294

L .323

Thus, only 32 percent of those persons whose test results are positive actually have

the disease. Many students are often surprised at this result (they expect the per-

centage to be much higher, since the blood test seems to be a good one), so it is

probably worthwhile to present a second argument that, although less rigorous than

the preceding one, is probably more revealing. We now do so.

Since .5 percent of the population actually has the disease, it follows that, on the

average, 1 person out of every 200 tested will have it. The test will correctly conﬁrm

that this person has the disease with probability .95. Thus, on the average, out of every

200 persons tested, the test will correctly conﬁrm that .95 person has the disease.

On the other hand, however, out of the (on the average) 199 healthy people, the

test will incorrectly state that (199)(.01) of these people have the disease. Hence, for

every .95 diseased person that the test correctly states is ill, there are (on the average)

(199)(.01) healthy persons that the test incorrectly states are ill. Thus, the proportion

of time that the test result is correct when it states that a person is ill is

.95

.95 + (199)(.01)

294

L .323

Equation (3.1) is also useful when one has to reassess one’s personal probabilities

in the light of additional information. For instance, consider the examples that follow.

EXAMPLE 3e

Consider a medical practitioner pondering the following dilemma: “If I’m at least 80

percent certain that my patient has this disease, then I always recommend surgery,

whereas if I’m not quite as certain, then I recommend additional tests that are expen-

sive and sometimes painful. Now, initially I was only 60 percent certain that Jones

had the disease, so I ordered the series A test, which always gives a positive result

when the patient has the disease and almost never does when he is healthy. The test

result was positive, and I was all set to recommend surgery when Jones informed me,

for the ﬁrst time, that he was diabetic. This information complicates matters because,

although it doesn’t change my original 60 percent estimate of his chances of having

the disease in question, it does affect the interpretation of the results of the A test.

This is so because the A test, while never yielding a positive result when the patient

is healthy, does unfortunately yield a positive result 30 percent of the time in the case

of diabetic patients who are not suffering from the disease. Now what do I do? More

tests or immediate surgery?”

Solution. In order to decide whether or not to recommend surgery, the doctor should

ﬁrst compute her updated probability that Jones has the disease given that the A test

Section 3.3 Bayes’s Formula 69

result was positive. Let D denote the event that Jones has the disease and E the event

that the A test result is positive. The desired conditional probability is then

P(D|E) =

P(DE)

P(E)

P(D)P(E|D)

P(E|D)P(D) + P(E|D

)P(D

)

(.6)1

1(.6) + (.3)(.4)

= .833

Note that we have computed the probability of a positive test result by conditioning

on whether or not Jones has the disease and then using the fact that, because Jones is

a diabetic, his conditional probability of a positive result given that he does not have

the disease, P(E|D

), equals .3. Hence, as the doctor should now be over 80 percent

certain that Jones has the disease, she should recommend surgery. .

EXAMPLE 3f

At a certain stage of a criminal investigation, the inspector in charge is 60 percent con-

vinced of the guilt of a certain suspect. Suppose, however, that a new piece of evidence

which shows that the criminal has a certain characteristic (such as left-handedness,

baldness, or brown hair) is uncovered. If 20 percent of the population possesses this

characteristic, how certain of the guilt of the suspect should the inspector now be if it

turns out that the suspect has the characteristic?

Solution. Letting G denote the event that the suspect is guilty and C the event that

he possesses the characteristic of the criminal, we have

P(G|C) =

P(GC)

P(C)

P(C|G)P(G)

P(C|G)P(G) + P(C|G

)P(G

)

1(.6)

1(.6) + (.2)(.4)

L .882

where we have supposed that the probability of the suspect having the characteristic

if he is, in fact, innocent is equal to .2, the proportion of the population possessing the

characteristic. .

EXAMPLE 3g

In the world bridge championships held in Buenos Aires in May 1965, the famous

British bridge partnership of Terrence Reese and Boris Schapiro was accused of

cheating by using a system of ﬁnger signals that could indicate the number of hearts

held by the players. Reese and Schapiro denied the accusation, and eventually a hear-

ing was held by the British bridge league. The hearing was in the form of a legal

proceeding with prosecution and defense teams, both having the power to call and

cross-examine witnesses. During the course of the proceeding, the prosecutor exam-

ined speciﬁc hands played by Reese and Schapiro and claimed that their playing these

70 Chapter 3 Conditional Probability and Independence

hands was consistent with the hypothesis that they were guilty of having illicit knowl-

edge of the heart suit. At this point, the defense attorney pointed out that their play

of these hands was also perfectly consistent with their standard line of play. How-

ever, the prosecution then argued that, as long as their play was consistent with the

hypothesis of guilt, it must be counted as evidence toward that hypothesis. What do

you think of the reasoning of the prosecution?

Solution. The problem is basically one of determining how the introduction of new

evidence (in the preceding example, the playing of the hands) affects the probability

of a particular hypothesis. If we let H denote a particular hypothesis (such as the

hypothesis that Reese and Schapiro are guilty) and E the new evidence, then

P(H|E) =

P(HE)

P(E)

P(E|H)P(H)

P(E|H)P(H) + P(E|H

)[1 − P (H)]

(3.2)

where P(H) is our evaluation of the likelihood of the hypothesis before the intro-

duction of the new evidence. The new evidence will be in support of the hypothesis

whenever it makes the hypothesis more likely—that is, whenever P(H|E) Ú P(H).

From Equation (3.2), this will be the case whenever

P(E|H) Ú P(E|H)P(H) + P(E|H

)[1 − P(H)]

or, equivalently, whenever

P(E|H) Ú P(E|H

)

In other words, any new evidence can be considered to be in support of a particular

hypothesis only if its occurrence is more likely when the hypothesis is true than when

it is false. In fact, the new probability of the hypothesis depends on its initial proba-

bility and the ratio of these conditional probabilities, since, from Equation (3.2),

P(H|E) =

P(H)

P(H) + [1 − P(H)]

P(E|H

)

P(E|H)

Hence, in the problem under consideration, the play of the cards can be con-

sidered to support the hypothesis of guilt only if such play would have been more

likely if the partnership were cheating than if they were not. As the prosecutor never

made this claim, his assertion that the evidence is in support of the guilt hypothesis is

invalid. .

When the author of this text drinks iced tea at a coffee shop, he asks for a glass of

water along with the (same-sized) glass of tea. As he drinks the tea, he continuously

reﬁlls the tea glass with water. Assuming a perfect mixing of water and tea, he won-

dered about the probability that his ﬁnal gulp would be tea. This wonderment led to

part (a) of the following problem and to a very interesting answer.

EXAMPLE 3h

Urn 1 initially has n red molecules and urn 2 has n blue molecules. Molecules are

randomly removed from urn 1 in the following manner: After each removal from

urn 1, a molecule is taken from urn 2 (if urn 2 has any molecules) and placed in urn 1.

The process continues until all the molecules have been removed. (Thus, there are

2n removals in all.)

Section 3.3 Bayes’s Formula 71

(a) Find P(R), where R is the event that the ﬁnal molecule removed from urn 1

is red.

(b) Repeat the problem when urn 1 initially has r

red molecules and b

blue

molecules and urn 2 initially has r

red molecules and b

blue molecules.

Solution. (a) Focus attention on any particular red molecule, and let F be the event

that this molecule is the ﬁnal one selected. Now, in order for F to occur, the molecule

in question must still be in the urn after the ﬁrst n molecules have been removed (at

which time urn 2 is empty). So, letting N

be the event that this molecule is not the ith

molecule to be removed, we have

P(F) = P(N

···N

= P(N

)P(N

) ···P(N

···N

n−1

)P(F|N

···N

)



1 −



···



1 −



where the preceding formula uses the fact that the conditional probability that the

molecule under consideration is the ﬁnal molecule to be removed, given that it is still

in urn 1 when only n molecules remain, is, by symmetry, 1/n.

Therefore, if we number the n red molecules and let R

be the event that red

molecule number j is the ﬁnal molecule removed, then it follows from the preceding

formula that

P(R

) =



1 −



Because the events R

are mutually exclusive, we obtain

P(R) = P

⎛

⎜

⎝



j=1

⎞

⎟

⎠



j=1

P(R

) = (1 −

)

L e

−1

(b) Suppose now that urn i initially has r

red and b

blue molecules, for i = 1, 2. To

ﬁnd P(R), the probability that the ﬁnal molecule removed is red, focus attention on

any molecule that is initially in urn 1. As in part (a), it follows that the probability

that this molecule is the ﬁnal one removed is

p =



1 −

+ b



+ b

That is,



1 −



is the probability that the molecule under consideration

is still in urn 1 when urn 2 becomes empty, and

is the conditional probability,

given the preceding event, that the molecule under consideration is the ﬁnal molecule

removed. Hence, if we let O be the event that the last molecule removed is one of the

molecules originally in urn 1, then

P(O) = (r

+ b

)p =



1 −

+ b



To determine P(R), we condition on whether O occurs, to obtain

P(R) = P(R|O)P(O) + P(R|O

)P(O

)

+ b



1 −

+ b



+ b



1 −



1 −

+ b





72 Chapter 3 Conditional Probability and Independence

If r

+ b

= r

+ b

= n, so that both urns initially have n molecules, then, when n

is large,

P(L) L

+ b

−1

+ b

(1 − e

−1

)

The change in the probability of a hypothesis when new evidence is introduced can

be expressed compactly in terms of the change in the odds of that hypothesis, where

the concept of odds is deﬁned as follows.

Deﬁnition

The odds of an event A are deﬁned by

P(A)

P(A

)

P(A)

1 − P(A)

That is, the odds of an event A tell how much more likely it is that the event A occurs

than it is that it does not occur. For instance, if P(A) =

,thenP(A) = 2P(A

),so

the odds are 2. If the odds are equal to α, then it is common to say that the odds are

“α to 1” in favor of the hypothesis.

Consider now a hypothesis H that is true with probability P (H), and suppose that

new evidence E is introduced. Then the conditional probabilities, given the evidence

E,thatH is true and that H is not true are respectively given by

P(H|E) =

P(E|H)P(H)

P(E)

P(H

|E) =

P(E|H

)P(H

)

P(E)

Therefore, the new odds after the evidence E has been introduced are

P(H|E)

P(H

|E)

P(H)

P(H

)

P(E|H)

P(E|H

)

(3.3)

That is, the new value of the odds of H is the old value, multiplied by the ratio of the

conditional probability of the new evidence given that H is true to the conditional

probability given that H is not true. Thus, Equation (3.3) veriﬁes the result of Exam-

ple 3f, since the odds, and thus the probability of H, increase whenever the new evi-

dence is more likely when H is true than when it is false. Similarly, the odds decrease

whenever the new evidence is more likely when H is false than when it is true.

EXAMPLE 3i

An urn contains two type A coins and one type B coin. When a type A coin is ﬂipped,

it comes up heads with probability 1/4, whereas when a type B coin is ﬂipped, it comes

up heads with probability 3/4. A coin is randomly chosen from the urn and ﬂipped.

Given that the ﬂip landed on heads, what is the probability that it was a type A coin?

Solution. Let A be the event that a type A coin was ﬂipped, and let B = A

be the

event that a type B coin was ﬂipped. We want P(A|heads), where heads is the event

that the ﬂip landed on heads. From Equation (3.3), we see that

P(A|heads)

P(A

|heads)

P(A)

P(B)

P(heads|A)

P(heads|B)

2/3

1/3

1/4

3/4

= 2/3

Section 3.3 Bayes’s Formula 73

Hence, the odds are 2/3 : 1, or, equivalently, the probability is 2/5 that a type A coin

was ﬂipped. .

Equation (3.1) may be generalized as follows: Suppose that F

, F

, ..., F

are mutu-

ally exclusive events such that



i=1

= S

In other words, exactly one of the events F

, F

, ..., F

must occur. By writing

E =



i=1

and using the fact that the events EF

, i = 1, ..., n are mutually exclusive, we obtain

P(E) =



i=1

P(EF

)



i=1

P(E|F

)P(F

) (3.4)

Thus, Equation (3.4) shows how, for given events F

, F

, ..., F

, of which one and

only one must occur, we can compute P(E) by ﬁrst conditioning on which one of

the F

occurs. That is, Equation (3.4) states that P(E) is equal to a weighted average

of P(E|F

), each term being weighted by the probability of the event on which it is

conditioned.

EXAMPLE 3j

In Example 5j of Chapter 2, we considered the probability that, for a randomly shuf-

ﬂed deck, the card following the ﬁrst ace is some speciﬁed card, and we gave a combi-

natorial argument to show that this probability is

. Here is a probabilistic argument

based on conditioning: Let E be the event that the card following the ﬁrst ace is some

speciﬁed card, say, card x. To compute P(E), we ignore card x and condition on the

relative ordering of the other 51 cards in the deck. Letting O be the ordering gives

P(E) =



P(E|O)P(O)

Now, given O, there are 52 possible orderings of the cards, corresponding to hav-

ing card x being the ith card in the deck, i = 1,...,52. But because all 52! possi-

ble orderings were initially equally likely, it follows that, conditional on O, each

of the 52 remaining possible orderings is equally likely. Because card x will follow

the ﬁrst ace for only one of these orderings, we have P(E|O) = 1/52, implying that

P(E) = 1/52. .

Again, let F

,...,F

be a set of mutually exclusive and exhaustive events (meaning

that exactly one of these events must occur).

Suppose now that E has occurred and we are interested in determining which one

of the F

also occurred. Then, by Equation (3.4), we have the following proposition.

74 Chapter 3 Conditional Probability and Independence

Proposition 3.1.

P(F

|E) =

P(EF

)

P(E)

P(E|F

)P(F

)



i=1

P(E|F

)P(F

)

(3.5)

Equation (3.5) is known as Bayes’s formula, after the English philosopher Thomas

Bayes. If we think of the events F

as being possible “hypotheses” about some sub-

ject matter, then Bayes’s formula may be interpreted as showing us how opinions

about these hypotheses held before the experiment was carried out [that is, the P(F

)]

should be modiﬁed by the evidence produced by the experiment.

EXAMPLE 3k

A plane is missing, and it is presumed that it was equally likely to have gone down

in any of 3 possible regions. Let 1 − β

, i = 1, 2, 3, denote the probability that the

plane will be found upon a search of the ith region when the plane is, in fact, in that

region. (The constants β

are called overlook probabilities, because they represent the

probability of overlooking the plane; they are generally attributable to the geograph-

ical and environmental conditions of the regions.) What is the conditional probability

that the plane is in the ith region given that a search of region 1 is unsuccessful?

Solution. Let R

, i = 1, 2, 3, be the event that the plane is in region i, and let E be

the event that a search of region 1 is unsuccessful. From Bayes’s formula, we obtain

P(R

|E) =

P(ER

)

P(E)

P(E|R

)P(R

)



i=1

P(E|R

)P(R

)

(β

)

(β

)

+ (1)

+ 2

For j = 2, 3,

P(R

|E) =

P(E|R

)P(R

)

P(E)

(1)

(β

)

+ 2

j = 2, 3

Note that the updated (that is, the conditional) probability that the plane is in

region j, given the information that a search of region 1 did not ﬁnd it, is greater than

Section 3.3 Bayes’s Formula 75

the initial probability that it was in region j when j Z 1 and is less than the initial prob-

ability when j = 1. This statement is certainly intuitive, since not ﬁnding the plane in

region 1 would seem to decrease its chance of being in that region and increase its

chance of being elsewhere. Further, the conditional probability that the plane is in

region 1 given an unsuccessful search of that region is an increasing function of the

overlook probability β

. This statement is also intuitive, since the larger β

is, the

more it is reasonable to attribute the unsuccessful search to “bad luck” as opposed

to the plane’s not being there. Similarly, P(R

|E), j Z 1, is a decreasing function

of β

. .

The next example has often been used by unscrupulous probability students to win

money from their less enlightened friends.

EXAMPLE 3l

Suppose that we have 3 cards that are identical in form, except that both sides of the

ﬁrst card are colored red, both sides of the second card are colored black, and one

side of the third card is colored red and the other side black. The 3 cards are mixed

up in a hat, and 1 card is randomly selected and put down on the ground. If the upper

side of the chosen card is colored red, what is the probability that the other side is

colored black?

Solution. Let RR, BB,andRB denote, respectively, the events that the chosen card

is all red, all black, or the red–black card. Also, let R be the event that the upturned

side of the chosen card is red. Then the desired probability is obtained by

P(RB|R) =

P(RB ∩ R)

P(R)

P(R|RB)P(RB)

P(R|RR)P(RR) + P(R|RB)P(RB) + P(R|BB)P(BB)







(1)











+ 0





Hence, the answer is

. Some students guess

as the answer by incorrectly reasoning

that, given that a red side appears, there are two equally likely possibilities: that the

card is the all-red card or the red–black card. Their mistake, however, is in assuming

that these two possibilities are equally likely. For, if we think of each card as consist-

ing of two distinct sides, then we see that there are 6 equally likely outcomes of the

experiment—namely, R

, R

, B

, R

, B

—where the outcome is R

if the ﬁrst side

of the all-red card is turned face up, R

if the second side of the all-red card is turned

face up, R

if the red side of the red–black card is turned face up, and so on. Since the

other side of the upturned red side will be black only if the outcome is R

, we see that

the desired probability is the conditional probability of R

given that either R

or R

occurred, which obviously equals

. .

EXAMPLE 3m

A new couple, known to have two children, has just moved into town. Suppose that

the mother is encountered walking with one of her children. If this child is a girl, what

is the probability that both children are girls?

76 Chapter 3 Conditional Probability and Independence

Solution. Let us start by deﬁning the following events:

: the ﬁrst (that is, the oldest) child is a girl.

: the second child is a girl.

G: the child seen with the mother is a girl.

Also, let B

, B

,andB denote similar events, except that “girl” is replaced by “boy.”

Now, the desired probability is P(G

|G), which can be expressed as follows:

P(G

|G) =

P(G

P(G)

P(G

)

P(G)

Also,

P(G) = P(G|G

)P(G

) + P (G|G

)P(G

)

+ P(G|B

)P(B

) + P(G|B

)P(B

)

= P(G

) + P(G|G

)P(G

) + P (G|B

)P(B

)

where the ﬁnal equation used the results P(G|G

) = 1andP(G|B

) = 0. If we

now make the usual assumption that all 4 gender possibilities are equally likely, then

we see that

P(G

|G) =

+ P(G|G

)/4 + P (G|B

)/4

1 + P(G|G

) + P (G|B

)

Thus, the answer depends on whatever assumptions we want to make about the con-

ditional probabilities that the child seen with the mother is a girl given the event G

and that the child seen with the mother is a girl given the event G

. For instance, if

we want to assume, on the one hand, that, independently of the genders of the chil-

dren, the child walking with the mother is the elder child with some probability p,

then it would follow that

P(G|G

) = p = 1 − P(G|B

)

implying under this scenario that

P(G

|G) =

If, on the other hand, we were to assume that if the children are of different genders,

then the mother would choose to walk with the girl with probability q, independently

of the birth order of the children, then we would have

P(G|G

) = P(G|B

) = q

implying that

P(G

|G) =

1 + 2q

For instance, if we took q = 1, meaning that the mother would always choose to walk

with a daughter, then the conditional probability the she has two daughters would be

Section 3.3 Bayes’s Formula 77

, which is in accord with Example 2b because seeing the mother with a daughter is

now equivalent to the event that she has at least one daughter.

Hence, as stated, the problem is incapable of solution. Indeed, even when the usual

assumption about equally likely gender probabilities is made, we still need to make

additional assumptions before a solution can be given. This is because the sample

space of the experiment consists of vectors of the form s

, s

, i, where s

is the gen-

der of the older child, s

is the gender of the younger child, and i identiﬁes the birth

order of the child seen with the mother. As a result, to specify the probabilities of

the events of the sample space, it is not enough to make assumptions only about

the genders of the children; it is also necessary to assume something about the con-

ditional probabilities as to which child is with the mother given the genders of the

children. .

EXAMPLE 3n

A bin contains 3 different types of disposable ﬂashlights. The probability that a type 1

ﬂashlight will give over 100 hours of use is .7, with the corresponding probabilities for

type 2 and type 3 ﬂashlights being .4 and .3, respectively. Suppose that 20 percent of

the ﬂashlights in the bin are type 1, 30 percent are type 2, and 50 percent are type 3.

(a) What is the probability that a randomly chosen ﬂashlight will give more than

100 hours of use?

(b) Given that a ﬂashlight lasted over 100 hours, what is the conditional probability

that it was a type j ﬂashlight, j = 1, 2, 3?

Solution. (a) Let A denote the event that the ﬂashlight chosen will give over 100

hours of use, and let F

be the event that a type j ﬂashlight is chosen, j = 1, 2, 3. To

compute P(A), we condition on the type of the ﬂashlight, to obtain

P(A) = P(A|F

)P(F

) + P (A|F

)P(F

) + P (A|F

)P(F

)

= (.7)(.2) + (.4)(.3) + (.3)(.5) = .41

There is a 41 percent chance that the ﬂashlight will last for over 100 hours.

(b) The probability is obtained by using Bayes’s formula:

P(F

|A) =

P(AF

)

P(A)

P(A|F

)P(F

)

.41

Thus,

P(F

|A) = (.7)(.2)/.41 = 14/41

P(F

|A) = (.4)(.3)/.41 = 12/41

P(F

|A) = (.3)(.5)/.41 = 15/41

For instance, whereas the initial probability that a type 1 ﬂashlight is chosen is only

.2, the information that the ﬂashlight has lasted over 100 hours raises the probability

of this event to 14/41 L .341. .

EXAMPLE 3o

A crime has been committed by a solitary individual, who left some DNA at the

scene of the crime. Forensic scientists who studied the recovered DNA noted that

78 Chapter 3 Conditional Probability and Independence

only ﬁve strands could be identiﬁed and that each innocent person, independently,

would have a probability of 10

−5

of having his or her DNA match on all ﬁve strands.

The district attorney supposes that the perpetrator of the crime could be any of the

one million residents of the town. Ten thousand of these residents have been released

from prison within the past 10 years; consequently, a sample of their DNA is on ﬁle.

Before any checking of the DNA ﬁle, the district attorney feels that each of the ten

thousand ex-criminals has probability α of being guilty of the new crime, while each

of the remaining 990,000 residents has probability β, where α = cβ. (That is, the

district attorney supposes that each recently released convict is c times as likely to

be the crime’s perpetrator as is each town member who is not a recently released

convict.) When the DNA that is analyzed is compared against the database of the

ten thousand ex-convicts, it turns out that A. J. Jones is the only one whose DNA

matches the proﬁle. Assuming that the district attorney’s estimate of the relationship

between α and β is accurate, what is the probability that A. J. is guilty?

Solution. To begin, note that, because probabilities must sum to 1, we have

1 = 10,000α + 990,000β = (10,000c + 990,000)β

Thus,

β =

10,000c + 990,000

, α =

10,000c + 990,000

Now, let G be the event that A. J. is guilty, and let M denote the event that A. J. is

the only one of the ten thousand on ﬁle to have a match. Then

P(G|M) =

P(GM)

P(M)

P(G)P(M|G)

P(M|G)P(G) + P(M|G

)P(G

)

On the one hand, if A. J. is guilty, then he will be the only one to have a DNA match

if none of the others on ﬁle have a match. Therefore,

P(M|G) = (1 − 10

−5

)

9999

On the other hand, if A. J. is innocent, then in order for him to be the only match, his

DNA must match (which will occur with probability 10

−5

), all others in the database

must be innocent, and none of these others can have a match. Now, given that A. J.

is innocent, the conditional probability that all the others in the database are also

innocent is

P(all others innocent|AJ innocent) =

P(all in database innocent)

P(AJ innocent)

1 − 10,000α

1 − α

Also, the conditional probability, given their innocence, that none of the others in the

database will have a match is (1 − 10

−5

)

9999

. Therefore,

P(M|G

) = 10

−5



1 − 10,000α

1 − α



(1 − 10

−5

)

9999

Section 3.4 Independent Events 79

Because P(G) = α, the preceding formula gives

P(G|M) =

α + 10

−5

(1 − 10,000α)

.9 +

−5

Thus, if the district attorney’s initial feelings were that an arbitrary ex-convict was

100 times more likely to have committed the crime than was a nonconvict (that is,

c = 100), then α =

19,900

and

P(G|M) =

1.099

L 0.9099

If the district attorney initially felt that the appropriate ratio was c = 10, then α =

109, 000

and

P(G|M) =

1.99

L 0.5025

If the district attorney initially felt that the criminal was equally likely to be any of

the members of the town (c = 1),thenα = 10

−6

and

P(G|M) =

10.9

L 0.0917

Thus, the probability ranges from approximately 9 percent when the district attor-

ney’s initial assumption is that all the members of the population have the same

chance of being the perpetrator to approximately 91 percent when she assumes that

each ex-convict is 100 times more likely to be the criminal than is a speciﬁed townsper-

son who is not an ex-convict. .

3.4 INDEPENDENT EVENTS

The previous examples of this chapter show that P(E|F), the conditional probabil-

ity of E given F, is not generally equal to P(E), the unconditional probability of E.

In other words, knowing that F has occurred generally changes the chances of E’s

occurrence. In the special cases where P(E|F) does in fact equal P(E), we say that E

is independent of F. That is, E is independent of F if knowledge that F has occurred

does not change the probability that E occurs.

Since P(E|F) = P(EF)/P(F), it follows that E is independent of F if

P(EF) = P(E)P(F) (4.1)

The fact that Equation (4.1) is symmetric in E

and F shows that whenever E is inde-

pendent of F, F is also independent of E. We thus have the following deﬁnition.

Deﬁnition

Two events E and F are said to be independent if Equation (4.1) holds.

Two events E and F that are not independent are said to be dependent.

EXAMPLE 4a

A card is selected at random from an ordinary deck of 52 playing cards. If E is the

event that the selected card is an ace and F is the event that it is a spade, then E

80 Chapter 3 Conditional Probability and Independence

and F are independent. This follows because P(EF) =

, whereas P(E) =

and

P(F) =

. .

EXAMPLE 4b

Two coins are ﬂipped, and all 4 outcomes are assumed to be equally likely. If E is

the event that the ﬁrst coin lands on heads and F the event that the second lands

on tails, then E and F are independent, since P(EF) = P({(H, T)}) =

, whereas

P(E) = P({(H, H), (H, T)}) =

and P(F) = P({(H, T), (T, T)}) =

. .

EXAMPLE 4c

Suppose that we toss 2 fair dice. Let E

denote the event that the sum of the dice is 6

and F denote the event that the ﬁrst die equals 4. Then

P(E

F) = P({(4, 2)}) =

whereas

P(E

)P(F) =







216

Hence, E

and F are not independent. Intuitively, the reason for this is clear because

if we are interested in the possibility of throwing a 6 (with 2 dice), we shall be quite

happy if the ﬁrst die lands on 4 (or, indeed, on any of the numbers 1, 2, 3, 4, and 5),

for then we shall still have a possibility of getting a total of 6. If, however, the ﬁrst

die landed on 6, we would be unhappy because we would no longer have a chance of

getting a total of 6. In other words, our chance of getting a total of 6 depends on the

outcome of the ﬁrst die; thus, E

and F cannot be independent.

Now, suppose that we let E

be the event that the sum of the dice equals 7. Is E

independent of F? The answer is yes, since

P(E

F) = P({(4, 3)}) =

whereas

P(E

)P(F) =











We leave it for the reader to present the intuitive argument why the event that the

sum of the dice equals 7 is independent of the outcome on the ﬁrst die. .

EXAMPLE 4d

If we let E denote the event that the next president is a Republican and F the event

that there will be a major earthquake within the next year, then most people would

probably be willing to assume that E and F are independent. However, there would

probably be some controversy over whether it is reasonable to assume that E is inde-

pendent of G, where G is the event that there will be a recession within two years

after the election. .

We now show that if E is independent of F,thenE is also independent of F

Section 3.4 Independent Events 81

Proposition 4.1. If E and F are independent, then so are E and F

Proof. Assume that E and F are independent. Since E = EF ∪ EF

and EF and

are obviously mutually exclusive, we have

P(E) = P(EF) + P(EF

)

= P(E)P(F) + P(EF

)

or, equivalently,

P(EF

) = P(E)[1 − P(F)]

= P(E)P(F

)

and the result is proved.

Thus, if E is independent of F, then the probability of E’s occurrence is unchanged

by information as to whether or not F has occurred.

Suppose now that E is independent of F and is also independent of G.IsE then

necessarily independent of FG? The answer, somewhat surprisingly, is no, as the fol-

lowing example demonstrates.

EXAMPLE 4e

Two fair dice are thrown. Let E denote the event that the sum of the dice is 7. Let F

denote the event that the ﬁrst die equals 4 and G denote the event that the second

die equals 3. From Example 4c, we know that E is independent of F, and the same

reasoning as applied there shows that E is also independent of G; but clearly, E is not

independent of FG [since P(E|FG) = 1]. .

It would appear to follow from Example 4e that an appropriate deﬁnition of the

independence of three events E, F,andG would have to go further than merely

assuming that all of the





pairs of events are independent. We are thus led to the

following deﬁnition.

Deﬁnition

Three events E, F,andG are said to be independent if

P(EFG) = P(E)P(F)P(G)

P(EF) = P(E)P(F)

P(EG) = P(E)P(G)

P(FG) = P(F)P(G)

Note that if E, F,andG are independent, then E will be independent of any event

formed from F and G. For instance, E is independent of F ∪ G, since

P[E(F ∪ G)] = P(EF ∪ EG)

= P(EF) + P(EG) − P(EFG)

= P(E)P(F) + P(E)P(G) − P(E)P(FG)

= P(E)[P(F) + P(G) − P(FG)]

= P

(E)P(F ∪ G)

82 Chapter 3 Conditional Probability and Independence

Of course, we may also extend the deﬁnition of independence to more than three

events. The events E

, E

, ..., E

are said to be independent if, for every subset



, E



, ..., E



, r … n, of these events,

P(E



···E



) = P(E



)P(E



) ···P(E



)

Finally, we deﬁne an inﬁnite set of events to be independent if every ﬁnite subset of

those events is independent.

Sometimes, a probability experiment under consideration consists of performing a

sequence of subexperiments. For instance, if the experiment consists of continually

tossing a coin, we may think of each toss as being a subexperiment. In many cases, it

is reasonable to assume that the outcomes of any group of the subexperiments have

no effect on the probabilities of the outcomes of the other subexperiments. If such

is the case, we say that the subexperiments are independent. More formally, we say

that the subexperiments are independent if E

, E

, ..., E

, ... is necessarily an inde-

pendent sequence of events whenever E

is an event whose occurrence is completely

determined by the outcome of the ith subexperiment.

If each subexperiment has the same set of possible outcomes, then the subexperi-

ments are often called trials.

EXAMPLE 4f

An inﬁnite sequence of independent trials is to be performed. Each trial results in a

success with probability p and a failure with probability 1 − p. What is the probabil-

ity that

(a) at least 1 success occurs in the ﬁrst n trials;

(b) exactly k successes occur in the ﬁrst n trials;

Solution. In order to determine the probability of at least 1 success in the ﬁrst n trials,

it is easiest to compute ﬁrst the probability of the complementary event: that of no

successes in the ﬁrst n trials. If we let E

denote the event of a failure on the ith trial,

then the probability of no successes is, by independence,

P(E

···E

) = P(E

)P(E

) ···P(E

) = (1 − p)

Hence, the answer to part (a) is 1 − (1 − p)

To compute the answer to part (b), consider any particular sequence of the ﬁrst

n outcomes containing k successes and n − k failures. Each one of these sequences

will, by the assumed independence of trials, occur with probability p

(1 − p)

n−k

Since there are





such sequences (there are n!/k!(n − k)! permutations of k

successes and n − k failures), the desired probability in part (b) is

P{exactly k successes}=





(1 − p)

n−k

To answer part (c), we note that, by part (a), the probability of the ﬁrst n trials all

resulting in success is given by

P(E

···E

) = p

Section 3.4 Independent Events 83

Thus, using the continuity property of probabilities (Section 2.6), we see that the

desired probability is given by

⎛

⎝



i=1

⎞

⎠

= P

⎛

⎝

lim

n→q



i=1

⎞

⎠

= lim

n→q

⎛

⎝



i=1

⎞

⎠

= lim

0ifp < 1

1ifp = 1

EXAMPLE 4g

A system composed of n separate components is said to be a parallel system if it

functions when at least one of the components functions. (See Figure 3.2.) For such a

system, if component i, which is independent of the other components, functions with

probability p

, i = 1, ..., n, what is the probability that the system functions?

Solution. Let A

denote the event that component i functions. Then

P{system functions}=1 − P{system does not function}

= 1 − P{all components do not function}

= 1 − P

⎛

⎝



⎞

⎠

= 1 −



i=1

(1 − p

) by independence

FIGURE 3.2: Parallel System: Functions if Current Flows from A to B

EXAMPLE 4h

Independent trials consisting of rolling a pair of fair dice are performed. What is the

probability that an outcome of 5 appears before an outcome of 7 when the outcome

of a roll is the sum of the dice?

Solution. If we let E

denote the event that no 5 or 7 appears on the ﬁrst n − 1 trials

anda5appears on the nth trial, then the desired probability is

⎛

⎝



n=1

⎞

⎠



n=1

P(E

)

84 Chapter 3 Conditional Probability and Independence

Now, since P{5 on any trial}=

and P{7 on any trial}=

, we obtain, by the inde-

pendence of trials,

P(E

) =



1 −



n−1

Thus,

⎛

⎝



n=1

⎞

⎠



n=1





n−1

1 −

This result could also have been obtained by the use of conditional probabilities.

If we let E be the event that a 5 occurs before a 7, then we can obtain the desired

probability, P(E), by conditioning on the outcome of the ﬁrst trial, as follows: Let F

be the event that the ﬁrst trial results in a 5, let G be the event that it results in a 7, and

let H be the event that the ﬁrst trial results in neither a 5 nor a 7. Then, conditioning

on which one of these events occurs gives

P(E) = P(E|F)P(F) + P(E|G)P (G) + P (E|H)P(H)

However,

P(E|F) = 1

P(E|G) = 0

P(E|H) = P(E)

The ﬁrst two equalities are obvious. The third follows because if the ﬁrst outcome

results in neither a 5 nor a 7, then at that point the situation is exactly as it was when

the problem ﬁrst started—namely, the experimenter will continually roll a pair of fair

dice until either a 5 or 7 appears. Furthermore, the trials are independent; therefore,

the outcome of the ﬁrst trial will have no effect on subsequent rolls of the dice. Since

P(F) =

, P(G) =

,andP(H) =

, it follows that

P(E) =

+ P(E)

P(E) =

The reader should note that the answer is quite intuitive. That is, because a 5 occurs

on any roll with probability

and a 7 with probability

, it seems intuitive that the

odds that a 5 appears before a 7 should be 6 to 4 against. The probability should then

, as indeed it is.

The same argument shows that if E and F are mutually exclusive events of an

experiment, then, when independent trials of the experiment are performed, the

event E will occur before the event F with probability

P(E)

P(E) + P(F)

Section 3.4 Independent Events 85

EXAMPLE 4i

There are n types of coupons, and each new one collected is independently of type i

with probability p



i=1

= 1. Suppose k coupons are to be collected. If A

is the

event that there is at least one type i coupon among those collected, then, for i Z j,

ﬁnd

(a) P(A

)

(b) P(A

∪ A

)

Solution.

P(A

) = 1 − P(A

)

= 1 − P{no coupon is type i}

= 1 − (1 − p

)

where the preceding used that each coupon is, independently, not of type i with prob-

ability 1 − p

. Similarly,

P(A

∪ A

) = 1 − P(



∪ A

)



= 1 − P{no coupon is either type i or type j}

= 1 − (1 − p

− p

)

where the preceding used that each coupon is, independently, neither of type i nor

type j with probability 1 − p

− p

To determine P(A

), we will use the identity

P(A

∪ A

) = P(A

) + P(A

) − P (A

)

which, in conjunction with parts (a) and (b), yields

P(A

) = 1 − (1 − p

)

+ 1 − (1 − p

)

− [1 − (1 − p

− p

)

]

= 1 − (1 − p

)

− (1 − p

)

+ (1 − p

− p

)

Consequently,

P(A

) =

P(A

)

P(A

)

1 − (1 − p

)

− (1 − p

)

+ (1 − p

− p

)

1 − (1 − p

)

The next example presents a problem that occupies an honored place in the his-

tory of probability theory. This is the famous problem of the points. In general terms,

the problem is this: Two players put up stakes and play some game, with the stakes

to go to the winner of the game. An interruption requires them to stop before either

has won and when each has some sort of a “partial score.” How should the stakes be

divided?

This problem was posed to the French mathematician Blaise Pascal in 1654 by

the Chevalier de M

e, who was a professional gambler at that time. In attacking

the problem, Pascal introduced the important idea that the proportion of the prize

deserved by the competitors should depend on their respective probabilities of win-

ning if the game were to be continued at that point. Pascal worked out some special

cases and, more importantly, initiated a correspondence with the famous French-

man Pierre de Fermat, who had a great reputation as a mathematician. The resulting

exchange of letters not only led to a complete solution to the problem of the points,

86 Chapter 3 Conditional Probability and Independence

but also laid the framework for the solution to many other problems connected with

games of chance. This celebrated correspondence, dated by some as the birth date

of probability theory, was also important in stimulating interest in probability among

the mathematicians in Europe, for Pascal and Fermat were both recognized as being

among the foremost mathematicians of the time. For instance, within a short time of

their correspondence, the young Dutch mathematician Christiaan Huygens came to

Paris to discuss these problems and solutions, and interest and activity in this new

ﬁeld grew rapidly.

EXAMPLE 4j The problem of the points

Independent trials resulting in a success with probability p and a failure with proba-

bility 1 − p are performed. What is the probability that n successes occur before m

failures? If we think of A and B as playing a game such that A gains 1 point when a

success occurs and B gains 1 point when a failure occurs, then the desired probability

is the probability that A would win if the game were to be continued in a position

where A needed n and B needed m more points to win.

Solution. We shall present two solutions. The ﬁrst is due to Pascal and the second to

Fermat.

Let us denote by P

n,m

the probability that n successes occur before m failures. By

conditioning on the outcome of the ﬁrst trial, we obtain

n,m

= pP

n−1,m

+ (1 − p)P

n,m−1

n Ú 1, m Ú 1

(Why? Reason it out.) Using the obvious boundary conditions P

n,0

= 0, P

0, m

= 1, we

can solve these equations for P

n,m

. Rather than go through the tedious details, let us

instead consider Fermat’s solution.

Fermat argued that, in order for n successes to occur before m failures, it is nec-

essary and sufﬁcient that there be at least n successes in the ﬁrst m + n − 1 trials.

(Even if the game were to end before a total of m + n − 1 trials were completed, we

could still imagine that the necessary additional trials were performed.) This is true,

for if there are at least n successes in the ﬁrst m + n − 1 trials, there could be at

most m − 1 failures in those m + n − 1 trials; thus, n successes would occur before

m failures. If, however, there were fewer than n successes in the ﬁrst m + n − 1

trials, there would have to be at least m failures in that same number of trials; thus, n

successes would not occur before m failures.

Hence, since, as shown in Example 4f, the probability of exactly k successes in

m + n − 1 trials is



m + n − 1



(1 − p)

m+n−1−k

, it follows that the desired

probability of n successes before m failures is

n,m

m+n−1



k=n



m + n − 1



(1 − p)

m+n−1−k

Our next two examples deal with gambling problems, with the ﬁrst having a sur-

prisingly elegant analysis.

∗

EXAMPLE 4k

Suppose that initially there are r players, with player i having n

units, n

> 0, i =

1, ..., r. At each stage, two of the players are chosen to play a game, with the winner

∗

The remainder of this section should be considered optional.

Section 3.4 Independent Events 87

of the game receiving 1 unit from the loser. Any player whose fortune drops to 0 is

eliminated, and this continues until a single player has all n K



i=1

units, with

that player designated as the victor. Assuming that the results of successive games

are independent and that each game is equally likely to be won by either of its two

players, ﬁnd P

, the probability that player i is the victor?

Solution. To begin, suppose that there are n players, with each player initially having

1 unit. Consider player i. Each stage she plays will be equally likely to result in her

either winning or losing 1 unit, with the results from each stage being independent.

In addition, she will continue to play stages until her fortune becomes either 0 or n.

Because this is the same for all n players, it follows that each player has the same

chance of being the victor, implying that each player has probability 1/n of being the

victor. Now, suppose these n players are divided into r teams, with team i containing

players, i = 1, ..., r. Then the probability that the victor is a member of team i is

/n. But because

(a) team i initially has a total fortune of n

units, i = 1, ..., r,and

(b) each game played by members of different teams is equally likely to be won by

either player and results in the fortune of members of the winning team increas-

ing by 1 and the fortune of the members of the losing team decreasing by 1,

it is easy to see that the probability that the victor is from team i is exactly the prob-

ability we desire. Thus, P

= n

/n. Interestingly, our argument shows that this result

does not depend on how the players in each stage are chosen. .

In the gambler’s ruin problem, there are only 2 gamblers, but they are not assumed

to be of equal skill.

EXAMPLE 4l The gambler’s ruin problem

Two gamblers, A and B, bet on the outcomes of successive ﬂips of a coin. On each

ﬂip, if the coin comes up heads, A collects 1 unit from B, whereas if it comes up tails,

A pays 1 unit to B. They continue to do this until one of them runs out of money. If

it is assumed that the successive ﬂips of the coin are independent and each ﬂip results

in a head with probability p, what is the probability that A ends up with all the money

if he starts with i units and B starts with N − i units?

Solution. Let E denote the event that A ends up with all the money when he starts

with i and B starts with N − i, and to make clear the dependence on the initial fortune

of A, let P

= P(E). We shall obtain an expression for P(E) by conditioning on the

outcome of the ﬁrst ﬂip as follows: Let H denote the event that the ﬁrst ﬂip lands on

heads; then

= P(E) = P(E|H)P(H) + P(E|H

)P(H

)

= pP(E |H) + (1 − p)P(E|H

)

Now, given that the ﬁrst ﬂip lands on heads, the situation after the ﬁrst bet is that

A has i + 1 units and B has N − (i + 1). Since the successive ﬂips are assumed to

be independent with a common probability p of heads, it follows that, from that point

on, A’s probability of winning all the money is exactly the same as if the game were

just starting with A having an initial fortune of i + 1andB having an initial fortune

of N − (i + 1). Therefore,

P(E|H) = P

i+1

88 Chapter 3 Conditional Probability and Independence

and similarly,

P(E|H

) = P

i−1

Hence, letting q = 1 − p, we obtain

= pP

i+1

+ qP

i−1

i = 1, 2, ..., N − 1 (4.2)

By making use of the obvious boundary conditions P

= 0andP

= 1, we shall

now solve Equation (4.2). Since p + q = 1, these equations are equivalent to

+ qP

= pP

i+1

+ qP

i−1

i+1

− P

i−1

) i = 1, 2, ..., N − 1 (4.3)

Hence, since P

= 0, we obtain, from Equation (4.3),

− P

) =

− P

) =





. (4.4)

− P

i−1

− P

i−2

) =





i−1

− P

N−1

− P

N−2

) =





N−1

Adding the ﬁrst i − 1 equations of (4.4) yields

− P

= P











+ ··· +





i−1



⎧

⎪

⎨

⎪

⎩

1 − (q/p)

Z 1

= 1

Using the fact that P

= 1, we obtain

⎧

⎪

⎨

⎪

⎩

1 − (q/p)

if p Z

if p =

Section 3.4 Independent Events 89

Hence,

⎧

⎪

⎨

⎪

⎩

1 − (q/p)

if p Z

if p =

(4.5)

Let Q

denote the probability that B winds up with all the money when A starts

with i and B starts with N − i. Then, by symmetry to the situation described, and on

replacing p by q and i by N − i, it follows that

⎧

⎪

⎨

⎪

⎩

1 − (p/q)

N−i

1 − (p/q)

if q Z

N − i

if q =

Moreover, since q =

is equivalent to p =

, we have, when q Z

+ Q

1 − (q/p)

1 − (p/q)

N−i

1 − (p/q)

− p

(q/p)

− q

(p/q)

N−i

− p

N−i

− q

+ q

N−i

− q

= 1

This result also holds when p = q =

,so

+ Q

= 1

In words, this equation states that, with probability 1, either A or B will wind up

with all of the money; in other words, the probability that the game continues indeﬁ-

nitely with A’s fortune always being between 1 and N − 1 is zero. (The reader must

be careful because, a priori, there are three possible outcomes of this gambling game,

not two: Either A wins, or B wins, or the game goes on forever with nobody winning.

We have just shown that this last event has probability 0.)

As a numerical illustration of the preceding result, if A were to start with 5 units

and B with 10, then the probability of A’s winning would be

if p were

, whereas it

would jump to

1 −





1 −





L .87

if p were .6.

A special case of the gambler’s ruin problem, which is also known as the problem of

duration of play, was proposed to Huygens by Fermat in 1657. The version Huygens

proposed, which he himself solved, was that A and B have 12 coins each. They play

for these coins in a game with 3 dice as follows: Whenever 11 is thrown (by either—it

makes no difference who rolls the dice), A gives a coin to B. Whenever 14 is thrown,

B gives a coin to A. The person who ﬁrst wins all the coins wins the game. Since

90 Chapter 3 Conditional Probability and Independence

P{roll 11}=

216

and P{roll 14}=

216

, we see from Example 4h that, for A,thisis

just the gambler’s ruin problem with p =

, i = 12, and N = 24. The general form

of the gambler’s ruin problem was solved by the mathematician James Bernoulli and

published 8 years after his death in 1713.

For an application of the gambler’s ruin problem to drug testing, suppose that two

new drugs have been developed for treating a certain disease. Drug i has a cure rate

, i = 1, 2, in the sense that each patient treated with drug i will be cured with proba-

bility P

. These cure rates are, however, not known, and we are interested in ﬁnding a

method for deciding whether P

> P

or P

> P

. To decide on one of these alterna-

tives, consider the following test: Pairs of patients are to be treated sequentially, with

one member of the pair receiving drug 1 and the other drug 2. The results for each

pair are determined, and the testing stops when the cumulative number of cures from

one of the drugs exceeds the cumulative number of cures from the other by some

ﬁxed, predetermined number. More formally, let

1 if the patient in the jth pair that receives drug 1 is cured

0 otherwise

1 if the patient in the jth pair that receives drug 2 is cured

0 otherwise

For a predetermined positive integer M, the test stops after pair N, where N is the

ﬁrst value of n such that either

+ ··· + X

− (Y

+ ··· + Y

) = M

+ ··· + X

− (Y

+ ··· + Y

) =−M

In the former case, we assert that P

> P

and in the latter that P

> P

In order to help ascertain whether the foregoing is a good test, one thing we would

like to know is the probability that it leads to an incorrect decision. That is, for given

and P

, where P

> P

, what is the probability that the test will incorrectly assert

that P

> P

? To determine this probability, note that after each pair is checked,

the cumulative difference of cures using drug 1 versus drug 2 will go up by 1 with

probability P

(1 − P

)—since this is the probability that drug 1 leads to a cure and

drug 2 does not—or go down by 1 with probability (1 − P

, or remain the same

with probability P

+ (1 − P

)(1 − P

). Hence, if we consider only those pairs

in which the cumulative difference changes, then the difference will go up by 1 with

probability

P = P{up 1|up 1 or down 1}

(1 − P

)

(1 − P

) + (1 − P

and down by 1 with probability

1 − P =

(1 − P

)

(1 − P

) + (1 − P

Thus, the probability that the test will assert that P

> P

is equal to the probability

that a gambler who wins each (one-unit) bet with probability P will go down M before

going up M. But Equation (4.5), with i = M, N = 2M, shows that this probability is

given by

Section 3.4 Independent Events 91

P{test asserts that P

> P

}

= 1 −

1 −



1 − P



1 −



1 − P



= 1 −

1 +



1 − P



1 + γ

where

γ =

1 − P

(1 − P

)

(1 − P

)

For instance, if P

= .6andP

= .4, then the probability of an incorrect decision is

.017 when M = 5 and reduces to .0003 when M = 10. .

Suppose that we are presented with a set of elements and we want to determine

whether at least one member of the set has a certain property. We can attack this

question probabilistically by randomly choosing an element of the set in such a way

that each element has a positive probability of being selected. Then the original ques-

tion can be answered by a consideration of the probability that the randomly selected

element does not have the property of interest. If this probability is equal to 1, then

none of the elements of the set have the property; if it is less than 1, then at least one

element of the set has the property.

The ﬁnal example of this section illustrates this technique.

EXAMPLE 4m

The complete graph having n vertices is deﬁned to be a set of n points (called vertices)

in the plane and the





lines (called edges) connecting each pair of vertices. The

complete graph having 3 vertices is shown in Figure 3.3. Suppose now that each edge

in a complete graph having n vertices is to be colored either red or blue. For a ﬁxed

integer k, a question of interest is, Is there a way of coloring the edges so that no set

of k vertices has all of its





connecting edges the same color? It can be shown by

a probabilistic argument that if n is not too large, then the answer is yes.

FIGURE 3.3

92 Chapter 3 Conditional Probability and Independence

The argument runs as follows: Suppose that each edge is, independently, equally

likely to be colored either red or blue. That is, each edge is red with probability

Number the





sets of k vertices and deﬁne the events E

, i = 1, ...,





follows:

={all of the connecting edges of the ith set

of k vertices are the same color}

Now, since each of the





connecting edges of a set of k vertices is equally likely to

be either red or blue, it follows that the probability that they are all the same color is

P(E

) = 2





k(k−1)/2

Therefore, because

⎛

⎝



⎞

⎠

…



P(E

)(Boole’s inequality)

we ﬁnd that P







, the probability that there is a set of k vertices all of whose

connecting edges are similarly colored, satisﬁes

⎛

⎝



⎞

⎠

…







k(k−1)/2−1

Hence, if







k(k−1)/2−1

< 1

or, equivalently, if





< 2

k(k−1)/2−1

then the probability that at least one of the





sets of k vertices has all of its

connecting edges the same color is less than 1. Consequently, under the preceding

condition on n and k, it follows that there is a positive probability that no set of k

vertices has all of its connecting edges the same color. But this conclusion implies

that there is at least one way of coloring the edges for which no set of k vertices has

all of its connecting edges the same color. .

Remarks. (a) Whereas the preceding argument established a condition on n and

k that guarantees the existence of a coloring scheme satisfying the desired property,

it gives no information about how to obtain such a scheme (although one possibility

would be simply to choose the colors at random, check to see if the resulting coloring

satisﬁes the property, and repeat the procedure until it does).

Section 3.5 P(·|F ) Is a Probability 93

(b) The method of introducing probability into a problem whose statement is

purely deterministic has been called the probabilistic method.

†

Other examples of this

method are given in Theoretical Exercise 24 and Examples 2t and 2u of Chapter 7.

3.5 P(·|F) IS A PROBABILITY

Conditional probabilities satisfy all of the properties of ordinary probabilities, as is

proved by Proposition 5.1, which shows that P(E|F) satisﬁes the three axioms of a

probability.

Proposition 5.1.

(a) 0 … P(E|F) … 1.

(b) P(S|F) = 1.

, i = 1, 2, ..., are mutually exclusive events, then

⎛

⎝



⎞

⎠



P(E

|F)

Proof. To prove part (a), we must show that 0 … P(EF)/P(F) … 1. The left-side

inequality is obvious, whereas the right side follows because EF ( F, which implies

that P(EF) … P(F). Part (b) follows because

P(S|F) =

P(SF)

P(F)

= 1

Part (c) follows from

⎛

⎝



i=1

⎞

⎠

⎛

⎝





i=1



⎞

⎠

P(F)







P(F)

since

⎛

⎝



⎞

⎠

F =





P(E

P(F)



P(E

|F)

where the next-to-last equality follows because E

= Ø implies that

F = Ø.

†

See N. Alon, J. Spencer, and P. Erdos, The Probabilistic Method (New York: John Wiley &

Sons, Inc., 1992).

94 Chapter 3 Conditional Probability and Independence

If we deﬁne Q(E) = P(E|F), then, from Proposition 5.1, Q(E) may be regarded

as a probability function on the events of S. Hence, all of the propositions previously

proved for probabilities apply to Q(E). For instance, we have

Q(E

∪ E

) = Q(E

) + Q(E

) − Q(E

)

or, equivalently,

P(E

∪ E

|F) = P(E

|F) + P(E

|F) − P(E

|F)

Also, if we deﬁne the conditional probability Q(E

) by Q(E

) = Q(E

Q(E

), then, from Equation (3.1), we have

Q(E

) = Q(E

)Q(E

) + Q(E

)Q(E

) (5.1)

Since

Q(E

) =

Q(E

)

Q(E

)

P(E

|F)

P(E

|F)

P(E

P(F)

P(E

P(F)

= P(E

Equation (5.1) is equivalent to

P(E

|F) = P(E

F)P(E

|F) + P(E

F)P(E

|F)

EXAMPLE 5a

Consider Example 3a, which is concerned with an insurance company which believes

that people can be divided into two distinct classes: those who are accident prone

and those who are not. During any given year, an accident-prone person will have an

accident with probability .4, whereas the corresponding ﬁgure for a person who is not

prone to accidents is .2. What is the conditional probability that a new policyholder

will have an accident in his or her second year of policy ownership, given that the

policyholder has had an accident in the ﬁrst year?

Solution. If we let A be the event that the policyholder is accident prone and we let

, i = 1, 2, be the event that he or she has had an accident in the ith year, then the

desired probability P(A

) may be obtained by conditioning on whether or not the

policyholder is accident prone, as follows:

P(A

) = P(A

|AA

)P(A|A

) + P (A

)P(A

)

Now,

P(A|A

) =

P(A

)

P(A

|A)P(A)

P(A

)

However, P(A) is assumed to equal

, and it was shown in Example 3a that P(A

) =

.26. Hence,

P(A|A

) =

(.4)(.3)

.26

Section 3.5 P(·|F ) Is a Probability 95

Thus,

P(A

) = 1 − P(A|A

) =

Since P(A

|AA

) = .4andP(A

) = .2, it follows that

P(A

) = (.4)

+ (.2)

L .29

EXAMPLE 5b

A female chimp has given birth. It is not certain, however, which of two male chimps

is the father. Before any genetic analysis has been performed, it is felt that the

probability that male number 1 is the father is p and the probability that male number

2 is the father is 1 − p. DNA obtained from the mother, male number 1, and male

number 2 indicate that, on one speciﬁc location of the genome, the mother has the

gene pair (A, A), male number 1 has the gene pair (a, a), and male number 2 has the

gene pair (A, a). If a DNA test shows that the baby chimp has the gene pair (A, a),

what is the probability that male number 1 is the father?

Solution. Let all probabilities be conditional on the event that the mother has the

gene pair (A, A), male number 1 has the gene pair (a, a), and male number 2 has

the gene pair (A, a). Now, let M

be the event that male number i, i = 1, 2, is the

father, and let B

A,a

be the event that the baby chimp has the gene pair (A, a). Then

P(M

A,a

) is obtained as follows:

P(M

A,a

) =

P(M

A,a

)

P(B

A,a

)

P(B

A,a

)P(M

)

P(B

A,a

)P(M

) + P (B

A,a

)P(M

)

1 · p

1 · p + (1/2)(1 − p)

1 + p

Because

1+p

> p when p < 1, the information that the baby’s gene pair is (A, a)

increases the probability that male number 1 is the father. This result is intuitive

because it is more likely that the baby would have gene pair (A, a) if M

is true than

if M

is true (the respective conditional probabilities being 1 and 1/2). .

The next example deals with a problem in the theory of runs.

EXAMPLE 5c

Independent trials, each resulting in a success with probability p or a failure with

probability q = 1 − p, are performed. We are interested in computing the probability

that a run of n consecutive successes occurs before a run of m consecutive failures.

Solution. Let E be the event that a run of n consecutive successes occurs before a run

of m consecutive failures. To obtain P(E), we start by conditioning on the outcome of

the ﬁrst trial. That is, letting H denote the event that the ﬁrst trial results in a success,

we obtain

P(E) = pP(E|H) + qP(E |H

) (5.2)

96 Chapter 3 Conditional Probability and Independence

Now, given that the ﬁrst trial was successful, one way we can get a run of n successes

before a run of m failures would be to have the next n − 1 trials all result in successes.

So, let us condition on whether or not that occurs. That is, letting F be the event that

trials 2 through n all are successes, we obtain

P(E|H) = P(E|FH)P(F|H) + P (E |F

H)P(F

|H) (5.3)

On the one hand, clearly, P(E|FH) = 1; on the other hand, if the event F

H occurs,

then the ﬁrst trial would result in a success, but there would be a failure some time

during the next n − 1 trials. However, when this failure occurs, it would wipe out all

of the previous successes, and the situation would be exactly as if we started out with

a failure. Hence,

P(E|F

H) = P(E|H

)

Because the independence of trials implies that F and H are independent, and because

P(F) = p

n−1

, it follows from Equation (5.3) that

P(E|H) = p

n−1

+ (1 − p

n−1

)P(E|H

) (5.4)

We now obtain an expression for P(E|H

) in a similar manner. That is, we let G

denote the event that trials 2 through m are all failures. Then

P(E|H

) = P(E|GH

)P(G|H

) + P(E|G

)P(G

) (5.5)

Now, GH

is the event that the ﬁrst m trials all result in failures, so P(E|GH

) = 0.

Also, if G

occurs, then the ﬁrst trial is a failure, but there is at least one success

in the next m − 1 trials. Hence, since this success wipes out all previous failures, we

see that

P(E|G

) = P(E|H)

Thus, because P(G

) = P(G

) = 1 − q

m−1

, we obtain, from (5.5),

P(E|H

) = (1 − q

m−1

)P(E|H) (5.6)

Solving Equations (5.4) and (5.6) yields

P(E|H) =

n−1

+ q

m−1

− p

n−1

m−1

and

P(E|H

) =

(1 − q

m−1

n−1

+ q

m−1

− p

n−1

m−1

Thus,

P(E) = pP(E|H) + qP(E|H

)

+ qp

n−1

(1 − q

m−1

)

n−1

+ q

m−1

− p

n−1

m−1

n−1

(1 − q

)

n−1

+ q

m−1

− p

n−1

m−1

(5.7)

Section 3.5 P(·|F ) Is a Probability 97

It is interesting to note that, by the symmetry of the problem, the probability of

obtaining a run of m failures before a run of n successes would be given by Equa-

tion (5.7) with p and q interchanged and n and m interchanged. Hence, this probabil-

ity would equal

P{run of m failures before a run of n successes}

m−1

(1 − p

)

m−1

+ p

n−1

− q

m−1

n−1

(5.8)

Since Equations (5.7) and (5.8) sum to 1, it follows that, with probability 1, either a

run of n successes or a run of m failures will eventually occur.

As an example of Equation (5.7), we note that, in tossing a fair coin, the probability

that a run of 2 heads will precede a run of 3 tails is

. For 2 consecutive heads before

4 consecutive tails, the probability rises to

. .

In our next example, we return to the matching problem (Example 5m, Chapter 2)

and this time obtain a solution by using conditional probabilities.

EXAMPLE 5d

At a party, n men take off their hats. The hats are then mixed up, and each man

randomly selects one. We say that a match occurs if a man selects his own hat. What

is the probability of

(a) no matches?

(b) exactly k matches?

Solution. (a) Let E denote the event that no matches occur, and to make explicit the

dependence on n, write P

= P(E). We start by conditioning on whether or not the

ﬁrst man selects his own hat—call these events M and M

, respectively. Then

= P(E) = P(E|M)P(M) + P(E|M

)P(M

)

Clearly, P(E|M) = 0, so

= P(E|M

)

n − 1

(5.9)

Now, P(E|M

) is the probability of no matches when n − 1 men select from a set of

n − 1 hats that does not contain the hat of one of these men. This can happen in either

of two mutually exclusive ways: Either there are no matches and the extra man does

not select the extra hat (this being the hat of the man who chose ﬁrst), or there are

no matches and the extra man does select the extra hat. The probability of the ﬁrst of

these events is just P

n−1

, which is seen by regarding the extra hat as “belonging” to

the extra man. Because the second event has probability [1/(n − 1)]P

n−2

, we have

P(E|M

) = P

n−1

n − 1

n−2

Thus, from Equation (5.9),

n − 1

n−1

n−2

or, equivalently,

− P

n−1

=−

n−1

− P

n−2

) (5.10)

98 Chapter 3 Conditional Probability and Independence

However, since P

is the probability of no matches when n men select among their

own hats, we have

= 0 P

So, from Equation (5.10),

− P

=−

− P

)

=−

or P

−

− P

=−

− P

)

or P

−

and, in general,

−

− ··· +

(−1)

(b) To obtain the probability of exactly k matches, we consider any ﬁxed group of

k men. The probability that they, and only they, select their own hats is

n − 1

···

n − (k − 1)

n−k

(n − k)!

n−k

where P

n−k

is the conditional probability that the other n − k men, selecting among

their own hats, have no matches. Since there are





choices of a set of k men, the

desired probability of exactly k matches is

n−k

−

+ ··· +

(−1)

n−k

(n − k)!

An important concept in probability theory is that of the conditional independence

of events. We say that the events E

and E

are conditionally independent given F

if, given that F occurs, the conditional probability that E

occurs is unchanged by

information as to whether or not E

occurs. More formally, E

and E

are said to be

conditionally independent given F if

P(E

F) = P(E

|F) (5.11)

or, equivalently,

P(E

|F) = P(E

|F)P(E

|F) (5.12)

The notion of conditional independence can easily be extended to more than two

events, and this extension is left as an exercise.

The reader should note that the concept of conditional independence was implic-

itly employed in Example 5a, where it was assumed that the events that a policyholder

had an accident in his or her ith year, i = 1, 2, ..., were conditionally independent

given whether or not the person was accident prone. The following example, some-

times referred to as Laplace’s rule of succession, further illustrates the concept of

conditional independence.

EXAMPLE 5e Laplace’s rule of succession

There are k + 1 coins in a box. When ﬂipped, the ith coin will turn up heads with

probability i/k, i = 0, 1, ..., k. A coin is randomly selected from the box and is then

Section 3.5 P(·|F ) Is a Probability 99

repeatedly ﬂipped. If the ﬁrst n ﬂips all result in heads, what is the conditional prob-

ability that the (n + 1)st ﬂip will do likewise?

Solution. Let C

denote the event that the ith coin, i = 0, 1, ..., k, is initially selected;

let F

denote the event that the ﬁrst n ﬂips all result in heads; and let H be the event

that the (n + 1)st ﬂip is a head. The desired probability, P(H|F

), is now obtained as

follows:

P(H|F

) =



i=0

P(H|F

)P(C

)

Now, given that the ith coin is selected, it is reasonable to assume that the outcomes

will be conditionally independent, with each one resulting in a head with probability

i/k. Hence,

P(H|F

) = P(H|C

) =

Also,

P(C

) =

P(C

)

P(F

)

P(F

)P(C

)



j=0

P(F

)P(C

)

(i/k)

[1/(k + 1)]



j=0

(j/k)

[1/(k + 1)]

Thus,

P(H|F

) =



i=0

(i/k)

n+1



j=0

(j/k)

But if k is large, we can use the integral approximations



i=0





n+1

dx =

n + 2



j=0





dx =

n + 1

So, for k large,

P(H|F

) L

n + 1

n + 2

EXAMPLE 5f Updating information sequentially

Suppose there are n mutually exclusive and exhaustive possible hypotheses, with ini-

tial (sometimes referred to as prior ) probabilities P(H



i=1

P(H

) = 1. Now, if

information that the event E has occurred is received, then the conditional probabil-

ity that H

is the true hypothesis (sometimes referred to as the updated or posterior

probability of H

)is

P(H

|E) =

P(E|H

)P(H

)



P(E|H

)P(H

)

(5.13)

100 Chapter 3 Conditional Probability and Independence

Suppose now that we learn ﬁrst that E

has occurred and then that E

has occurred.

Then, given only the ﬁrst piece of information, the conditional probability that H

the true hypothesis is

P(H

) =

P(E

)P(H

)

P(E

)

P(E

)P(H

)



P(E

)P(H

)

whereas given both pieces of information, the conditional probability that H

is the

true hypothesis is P(H

), which can be computed by

P(H

) =

P(E

)P(H

)



P(E

)P(H

)

One might wonder, however, when one can compute P(H

) by using the

right side of Equation (5.13) with E = E

and with P(H

) replaced by P(H

j = 1, ..., n. That is, when is it legitimate to regard P(H

), j Ú 1, as the prior

probabilities and then use (5.13) to compute the posterior probabilities?

Solution. The answer is that the preceding is legitimate, provided that, for each j =

1, ..., n, the events E

and E

are conditionally independent, given H

. For if this is

the case, then

P(E

) = P(E

)P(E

), j = 1, ..., n

Therefore,

P(H

) =

P(E

)P(E

)P(H

)

P(E

)

P(E

)P(E

)

P(E

)

P(E

)P(H

)P(E

)

P(E

)

P(E

)P(H

)

Q(1, 2)

where Q(1, 2) =

P(E

)

P(E

)

. Since the preceding equation is valid for all i, we obtain,

upon summing,

1 =



i=1

P(H

) =



i=1

P(E

)P(H

)

Q(1, 2)

showing that

Q(1, 2) =



i=1

P(E

)P(H

)

and yielding the result

P(H

) =

P(E

)P(H

)



i=1

P(E

)P(H

)

For instance, suppose that one of two coins is chosen to be ﬂipped. Let H

be the event

that coin i, i = 1, 2, is chosen, and suppose that when coin i is ﬂipped, it lands on heads

with probability p

, i = 1, 2. Then the preceding equations show that, to sequentially

Summary 101

update the probability that coin 1 is the one being ﬂipped, given the results of the

previous ﬂips, all that must be saved after each new ﬂip is the conditional probability

that coin 1 is the coin being used. That is, it is not necessary to keep track of all earlier

results. .

SUMMARY

For events E and F, the conditional probability of E given that F has occurred is

denoted by P(E|F) andisdeﬁnedby

P(E|F) =

P(EF)

P(F)

The identity

P(E

···E

) = P(E

)P(E

) ···P(E

···E

n−1

)

is known as the multiplication rule of probability.

A valuable identity is

P(E) = P(E|F)P(F) + P(E|F

)P(F

)

which can be used to compute P(E) by “conditioning” on whether F occurs.

P(H)/P(H

) is called the odds of the event H. The identity

P(H|E)

P(H

|E)

P(H) P(E|H)

P(H

)P(E|H

)

shows that when new evidence E is obtained, the value of the odds of H becomes its

old value multiplied by the ratio of the conditional probability of the new evidence

when H is true to the conditional probability when H is not true.

Let F

, i = 1, ..., n, be mutually exclusive events whose union is the entire sample

space. The identity

P(F

|E) =

P(E|F

)P(F

)



i=1

P(E|F

)P(F

)

is known as Bayes’s formula. If the events F

, i = 1, ..., n, are competing hypotheses,

then Bayes’s formula shows how to compute the conditional probabilities of these

hypotheses when additional evidence E becomes available.

If P(EF) = P(E)P(F), then we say that the events E and F are independent. This

condition is equivalent to P(E|F) = P(E) and to P(F|E) = P(F). Thus, the events E

and F are independent if knowledge of the occurrence of one of them does not affect

the probability of the other.

The events E

, ..., E

are said to be independent if, for any subset E

, ..., E

of them,

P(E

···E

) = P(E

) ···P(E

)

For a ﬁxed event F, P(E|F) can be considered to be a probability function on the

events E of the sample space.

102 Chapter 3 Conditional Probability and Independence

PROBLEMS

3.1. Two fair dice are rolled. What is the conditional

probability that at least one lands on 6 given that

the dice land on different numbers?

3.2. If two fair dice are rolled, what is the conditional

probability that the ﬁrst one lands on 6 given that

thesumofthediceisi? Compute for all values of

i between 2 and 12.

3.3. Use Equation (2.1) to compute, in a hand of

bridge, the conditional probability that East has

3 spades given that North and South have a com-

bined total of 8 spades.

3.4. What is the probability that at least one of a pair of

fair dice lands on 6, given that the sum of the dice

is i, i = 2, 3, ..., 12?

3.5. An urn contains 6 white and 9 black balls. If 4 balls

are to be randomly selected without replacement,

what is the probability that the ﬁrst 2 selected are

white and the last 2 black?

3.6. Consider an urn containing 12 balls, of which 8

arewhite.Asampleofsize4istobedrawnwith

replacement (without replacement). What is the

conditional probability (in each case) that the ﬁrst

and third balls drawn will be white given that the

sample drawn contains exactly 3 white balls?

3.7. The king comes from a family of 2 children. What

is the probability that the other child is his sister?

3.8. A couple has 2 children. What is the probability

that both are girls if the older of the two is a girl?

3.9. Consider 3 urns. Urn A contains 2 white and 4 red

balls, urn B contains 8 white and 4 red balls, and

urn C contains 1 white and 3 red balls. If 1 ball is

selected from each urn, what is the probability that

the ball chosen from urn A was white given that

exactly 2 white balls were selected?

3.10. Three cards are randomly selected, without

replacement, from an ordinary deck of 52 playing

cards. Compute the conditional probability that

the ﬁrst card selected is a spade given that the sec-

ond and third cards are spades.

3.11. Two cards are randomly chosen without replace-

ment from an ordinary deck of 52 cards. Let B be

the event that both cards are aces, let A

be the

event that the ace of spades is chosen, and let A be

the event that at least one ace is chosen. Find

(a) P(B|A

)

(b) P(B|A)

3.12. A recent college graduate is planning to take the

ﬁrst three actuarial examinations in the coming

summer. She will take the ﬁrst actuarial exam in

June. If she passes that exam, then she will take

the second exam in July, and if she also passes that

one, then she will take the third exam in Septem-

ber. If she fails an exam, then she is not allowed

to take any others. The probability that she passes

the ﬁrst exam is .9. If she passes the ﬁrst exam, then

the conditional probability that she passes the sec-

ond one is .8, and if she passes both the ﬁrst and

the second exams, then the conditional probability

that she passes the third exam is .7.

(a) What is the probability that she passes all

three exams?

(b) Given that she did not pass all three exams,

what is the conditional probability that she

failed the second exam?

3.13. Suppose that an ordinary deck of 52 cards (which

contains 4 aces) is randomly divided into 4 hands

of 13 cards each. We are interested in determining

p, the probability that each hand has an ace. Let E

be the event that the ith hand has exactly one ace.

Determine p = P(E

) by using the multi-

plication rule.

3.14. An urn initially contains 5 white and 7 black balls.

Each time a ball is selected, its color is noted and

it is replaced in the urn along with 2 other balls of

the same color. Compute the probability that

(a) the ﬁrst 2 balls selected are black and the next

2 are white;

(b) of the ﬁrst 4 balls selected, exactly 2 are black.

3.15. An ectopic pregnancy is twice as likely to develop

when the pregnant woman is a smoker as it is when

she is a nonsmoker. If 32 percent of women of

childbearing age are smokers, what percentage of

women having ectopic pregnancies are smokers?

3.16. Ninety-eight percent of all babies survive delivery.

However, 15 percent of all births involve Cesarean

the baby survives 96 percent of the time. If a ran-

domly chosen pregnant woman does not have a

C section, what is the probability that her baby

survives?

3.17. In a certain community, 36 percent of the families

own a dog and 22 percent of the families that own

a dog also own a cat. In addition, 30 percent of the

families own a cat. What is

(a) the probability that a randomly selected fam-

ily owns both a dog and a cat?

(b) the conditional probability that a randomly

selected family owns a dog given that it owns

acat?

3.18. A total of 46 percent of the voters in a certain city

classify themselves as Independents, whereas 30

percent classify themselves as Liberals and 24 per-

cent say that they are Conservatives. In a recent

local election, 35 percent of the Independents, 62

percent of the Liberals, and 58 percent of the Con-

servatives voted. A voter is chosen at random.

Problems 103

Given that this person voted in the local election,

what is the probability that he or she is

(a) an Independent?

(b) a Liberal?

(d) What fraction of voters participated in the

local election?

3.19. A total of 48 percent of the women and 37 percent

of the men that took a certain “quit smoking” class

remained nonsmokers for at least one year after

completing the class. These people then attended

a success party at the end of a year. If 62 percent

of the original class was male,

(a) what percentage of those attending the party

were women?

(b) what percentage of the original class attended

the party?

3.20. Fifty-two percent of the students at a certain col-

lege are females. Five percent of the students in

this college are majoring in computer science. Two

percent of the students are women majoring in

computer science. If a student is selected at ran-

dom, ﬁnd the conditional probability that

(a) the student is female given that the student is

majoring in computer science;

(b) this student is majoring in computer science

given that the student is female.

3.21. A total of 500 married working couples were

polled about their annual salaries, with the follow-

ing information resulting:

Husband

Wife Less than More than

$25,000 $25,000

Less than $25,000 212 198

More than $25,000 36 54

For instance, in 36 of the couples, the wife earned

more and the husband earned less than $25,000. If

one of the couples is randomly chosen, what is

(a) the probability that the husband earns less

than $25,000?

(b) the conditional probability that the wife earns

more than $25,000 given that the husband

earns more than this amount?

more than $25,000 given that the husband

earns less than this amount?

3.22. A red die, a blue die, and a yellow die (all six

sided) are rolled. We are interested in the prob-

ability that the number appearing on the blue die

is less than that appearing on the yellow die, which

is less than that appearing on the red die. That is,

with B, Y,andR denoting, respectively, the num-

ber appearing on the blue, yellow, and red die, we

are interested in P(B < Y < R).

(a) What is the probability that no two of the dice

land on the same number?

(b) Given that no two of the dice land on the same

number, what is the conditional probability

that B < Y < R?

3.23. Urn I contains 2 white and 4 red balls, whereas urn

II contains 1 white and 1 red ball. A ball is ran-

domly chosen from urn I and put into urn II, and a

ball is then randomly selected from urn II. What is

(a) the probability that the ball selected from urn

II is white?

(b) the conditional probability that the trans-

ferred ball was white given that a white ball

is selected from urn II?

3.24. Each of 2 balls is painted either black or gold and

then placed in an urn. Suppose that each ball is col-

ored black with probability

and that these events

are independent.

(a) Suppose that you obtain information that the

gold paint has been used (and thus at least

one of the balls is painted gold). Compute

the conditional probability that both balls are

painted gold.

(b) Suppose now that the urn tips over and 1 ball

falls out. It is painted gold. What is the prob-

ability that both balls are gold in this case?

Explain.

3.25. The following method was proposed to estimate

the number of people over the age of 50 who reside

in a town of known population 100,000: “As you

walk along the streets, keep a running count of the

percentage of people you encounter who are over

50. Do this for a few days; then multiply the per-

centage you obtain by 100,000 to obtain the esti-

mate.” Comment on this method.

Hint:Letp denote the proportion of people in the

town who are over 50. Furthermore, let α

denote

the proportion of time that a person under the age

of 50 spends in the streets, and let α

be the cor-

responding value for those over 50. What quantity

does the method suggested estimate? When is the

estimate approximately equal to p?

3.26. Suppose that 5 percent of men and .25 percent

of women are color blind. A color-blind person

is chosen at random. What is the probability of

this person being male? Assume that there are an

equal number of males and females. What if the

population consisted of twice as many males as

females?

3.27. All the workers at a certain company drive to

work and park in the company’s lot. The company

104 Chapter 3 Conditional Probability and Independence

is interested in estimating the average number of

workers in a car. Which of the following methods

will enable the company to estimate this quantity?

Explain your answer.

1. Randomly choose n workers, ﬁnd out how

many were in the cars in which they were

driven, and take the average of the n values.

2. Randomly choose n cars in the lot, ﬁnd out how

many were driven in those cars, and take the

average of the n values.

3.28. Suppose that an ordinary deck of 52 cards is shuf-

ﬂed and the cards are then turned over one at a

time until the ﬁrst ace appears. Given that the ﬁrst

ace is the 20th card to appear, what is the condi-

tional probability that the card following it is the

(a) ace of spades?

(b) two of clubs?

3.29. There are 15 tennis balls in a box, of which 9 have

not previously been used. Three of the balls are

randomly chosen, played with, and then returned

to the box. Later, another 3 balls are randomly

chosen from the box. Find the probability that

none of these balls has ever been used.

3.30. Consider two boxes, one containing 1 black and 1

white marble, the other 2 b lack and 1 white mar-

ble. A box is selected at random, and a marble is

drawn from it at random. What is the probability

that the marble is black? What is the probability

that the ﬁrst box was the one selected given that

the marble is white?

3.31. Ms. Aquina has just had a biopsy on a possibly can-

cerous tumor. Not wanting to spoil a weekend fam-

ily event, she does not want to hear any bad news

in the next few days. But if she tells the doctor to

call only if the news is good, then if the doctor does

not call, Ms. Aquina can conclude that the news is

bad. So, being a student of probability, Ms. Aquina

instructs the doctor to ﬂip a coin. If it comes up

heads, the doctor is to call if the news is good and

not call if the news is bad. If the coin comes up

tails, the doctor is not to call. In this way, even if

the doctor doesn’t call, the news is not necessarily

bad. Let α be the probability that the tumor is can-

cerous; let β be the conditional probability that the

tumor is cancerous given that the doctor does not

call.

(a) Which should be larger, α or β?

(b) Find β in terms of α, and prove your answer

in part (a).

3.32. A family has j children with probability p

,where

= .1, p

= .25, p

= .35, p

= .3. A child

from this family is randomly chosen. Given that

this child is the eldest child in the family, ﬁnd the

conditional probability that the family has

(a) only 1 child;

(b) 4 children.

Redo (a) and (b) when the randomly selected child

is the youngest child of the family.

3.33. On rainy days, Joe is late to work with probability

.3; on nonrainy days, he is late with probability .1.

With probability .7, it will rain tomorrow.

(a) Find the probability that Joe is early tomor-

row.

(b) Given that Joe was early, what is the condi-

tional probability that it rained?

3.34. In Example 3f, suppose that the new evidence is

subject to different possible interpretations and in

fact shows only that it is 90 percent likely that the

criminal possesses the characteristic in question. In

this case, how likely would it be that the suspect is

guilty (assuming, as before, that he has the charac-

teristic)?

3.35. With probability .6, the present was hidden by

mom; with probability .4, it was hidden by dad.

When mom hides the present, she hides it upstairs

70 percent of the time and downstairs 30 percent

of the time. Dad is equally likely to hide it upstairs

or downstairs.

(a) What is the probability that the present is

upstairs?

(b) Given that it is downstairs, what is the proba-

bility it was hidden by dad?

3.36. Stores A, B,andC have 50, 75, and 100 employees,

respectively, and 50, 60, and 70 percent of them

respectively are women. Resignations are equally

likely among all employees, regardless of sex. One

woman employee resigns. What is the probability

that she works in store C?

3.37. (a) A gambler has a fair coin and a two-headed

coin in his pocket. He selects one of the coins

at random; when he ﬂips it, it shows heads.

What is the probability that it is the fair coin?

(b) Suppose that he ﬂips the same coin a second

time and, again, it shows heads. Now what is

the probability that it is the fair coin?

time and it shows tails. Now what is the prob-

ability that it is the fair coin?

3.38. Urn A has 5 white and 7 black balls. Urn B has

3 white and 12 black balls. We ﬂip a fair coin. If

the outcome is heads, then a ball from urn A is

selected, whereas if the outcome is tails, then a ball

from urn B is selected. Suppose that a white ball

is selected. What is the probability that the coin

landed tails?

3.39. In Example 3a, what is the probability that some-

one has an accident in the second year given that

he or she had no accidents in the ﬁrst year?

3.40. Consider a sample of size 3 drawn in the following

manner: We start with an urn containing 5 white

Problems 105

and 7 red balls. At each stage, a ball is drawn

and its color is noted. The ball is then returned

to the urn, along with an additional ball of the

same color. Find the probability that the sample

will contain exactly

(a) 0 white balls;

(b) 1 white ball;

(d) 2 white balls.

3.41. A deck of cards is shufﬂed and then divided into

two halves of 26 cards each. A card is drawn from

one of the halves; it turns out to be an ace. The ace

is then placed in the second half-deck. The half is

then shufﬂed, and a card is drawn from it. Com-

pute the probability that this drawn card is an ace.

Hint: Condition on whether or not the inter-

changed card is selected.

3.42. Three cooks, A, B,andC, bake a special kind of

cake, and with respective probabilities .02, .03, and

.05, it fails to rise. In the restaurant where they

work, A bakes 50 percent of these cakes, B 30 per-

cent, and C 20 percent. What proportion of “fail-

ures” is caused by A?

3.43. There are 3 coins in a box. One is a two-headed

coin, another is a fair coin, and the third is a biased

coin that comes up heads 75 percent of the time.

When one of the 3 coins is selected at random and

ﬂipped, it shows heads. What is the probability that

it was the two-headed coin?

3.44. Three prisoners are informed by their jailer that

one of them has been chosen at random to be

executed and the other two are to be freed. Pris-

oner A asks the jailer to tell him privately which of

his fellow prisoners will be set free, claiming that

there would be no harm in divulging this informa-

tion because he already knows that at least one of

the two will go free. The jailer refuses to answer

the question, pointing out that if A knew which

of his fellow prisoners were to be set free, then

his own probability of being executed would rise

from

because he would then be one of two

prisoners. What do you think of the jailer’s

reasoning?

3.45. Suppose we have 10 coins such that if the ith

coin is ﬂipped, heads will appear with probabil-

ity i/10, i = 1, 2, ..., 10. When one of the coins

is randomly selected and ﬂipped, it shows heads.

What is the conditional probability that it was the

ﬁfth coin?

3.46. In any given year, a male automobile policyholder

will make a claim with probability p

and a female

policyholder will make a claim with probability p

where p

Z p

. The fraction of the policyholders

that are male is α,0 <α<1. A policyholder is

randomly chosen. If A

denotes the event that this

policyholder will make a claim in year i, show that

P(A

)>P(A

)

Give an intuitive explanation of why the preceding

inequality is true.

3.47. An urn contains 5 white and 10 black balls. A fair

die is rolled and that number of balls is randomly

chosen from the urn. What is the probability that

all of the balls selected are white? What is the con-

ditional probability that the die landed on 3 if all

the balls selected are white?

3.48. Each of 2 cabinets identical in appearance has 2

drawers. Cabinet A contains a silver coin in each

drawer, and cabinet B contains a silver coin in

one of its drawers and a gold coin in the other.

A cabinet is randomly selected, one of its drawers

is opened, and a silver coin is found. What is the

probability that there is a silver coin in the other

drawer?

3.49. Prostate cancer is the most common type of can-

cer found in males. As an indicator of whether a

male has prostate cancer, doctors often perform

a test that measures the level of the prostate-

speciﬁc antigen (PSA) that is produced only by the

prostate gland. Although PSA levels are indica-

tive of cancer, the test is notoriously unreli-

able. Indeed, the probability that a noncancerous

man will have an elevated PSA level is approx-

imately .135, increasing to approximately .268 if

the man does have cancer. If, on the basis of other

factors, a physician is 70 percent certain that a

male has prostate cancer, what is the conditional

probability that he has the cancer given that

(a) the test indicated an elevated PSA level?

(b) the test did not indicate an elevated PSA

level?

Repeat the preceding calculation, this time assum-

ing that the physician initially believes that there

is a 30 percent chance that the man has prostate

cancer.

3.50. Suppose that an insurance company classiﬁes peo-

ple into one of three classes: good risks, average

risks, and bad risks. The company’s records indi-

cate that the probabilities that good-, average-, and

bad-risk persons will be involved in an accident

over a 1-year span are, respectively, .05, .15, and

.30. If 20 percent of the population is a good risk,

50 percent an average risk, and 30 percent a bad

risk, what proportion of people have accidents in

a ﬁxed year? If policyholder A had no accidents

in 1997, what is the probability that he or she is a

good or average risk?

3.51. A worker has asked her supervisor for a letter of

recommendation for a new job. She estimates that

there is an 80 percent chance that she will get the

106 Chapter 3 Conditional Probability and Independence

job if she receives a strong recommendation, a 40

percent chance if she receives a moderately good

recommendation, and a 10 percent chance if she

receives a weak recommendation. She further esti-

mates that the probabilities that the recommenda-

tion will be strong, moderate, and weak are .7, .2,

and .1, respectively.

(a) How certain is she that she will receive the

new job offer?

(b) Given that she does receive the offer, how

likely should she feel that she received a

strong recommendation? a moderate recom-

mendation? a weak recommendation?

how likely should she feel that she received a

strong recommendation? a moderate recom-

mendation? a weak recommendation?

3.52. A high school student is anxiously waiting to

receive mail telling her whether she has been

accepted to a certain college. She estimates that

the conditional probabilities of receiving notiﬁca-

tion on each day of next week, given that she is

accepted and that she is rejected, are as follows:

Day P(mail|accepted) P(mail|rejected)

Monday .15 .05

Tuesday .20 .10

Wednesday .25 .10

Thursday .15 .15

Friday .10 .20

She estimates that her probability of being

accepted is .6.

(a) What is the probability that she receives mail

on Monday?

(b) What is the conditional probability that she

received mail on Tuesday given that she does

not receive mail on Monday?

is the conditional probability that she will be

accepted?

(d) What is the conditional probability that she

will be accepted if mail comes on Thursday?

(e) What is the conditional probability that she

will be accepted if no mail arrives that week?

3.53. A parallel system functions whenever at least one

of its components works. Consider a parallel sys-

tem of n components, and suppose that each com-

ponent works independently with probability

Find the conditional probability that component 1

works given that the system is functioning.

3.54. If you had to construct a mathematical model for

events E and F, as described in parts (a) through

(e), would you assume that they were independent

events? Explain your reasoning.

(a) E is the event that a businesswoman has blue

eyes, and F is the event that her secretary has

blue eyes.

(b) E is the event that a professor owns a car,

and F is the event that he is listed in the tele-

phone book.

and F is the event that he weighs over 200

pounds.

(d) E is the event that a woman lives in the United

States, and F is the event that she lives in the

Western Hemisphere.

(e) E is the event that it will rain tomorrow, and

F is the event that it will rain the day after

tomorrow.

3.55. In a class, there are 4 freshman boys, 6 freshman

girls, and 6 sophomore boys. How many sopho-

more girls must be present if sex and class are to

be independent when a student is selected at ran-

dom?

3.56. Suppose that you continually collect coupons and

that there are m different types. Suppose also that

each time a new coupon is obtained, it is a type

i coupon with probability p

, i = 1, ..., m. Suppose

that you have just collected your nth coupon. What

is the probability that it is a new type?

Hint: Condition on the type of this coupon.

3.57. A simpliﬁed model for the movement of the price

of a stock supposes that on each day the stock’s

price either moves up 1 unit with probability p or

moves down 1 unit with probability 1 − p.The

changes on different days are assumed to be inde-

pendent.

(a) What is the probability that after 2 days the

stock will be at its original price?

(b) What is the probability that after 3 days the

stock’s price will have increased by 1 unit?

increased by 1 unit, what is the probability

that it went up on the ﬁrst day?

3.58. Suppose that we want to generate the outcome

of the ﬂip of a fair coin, but that all we have at

our disposal is a biased coin which lands on heads

with some unknown probability p that need not be

equal to

. Consider the following procedure for

accomplishing our task:

1. Flip the coin.

2. Flip the coin again.

3. If both ﬂips land on heads or both land on tails,

return to step 1.

4. Let the result of the last ﬂip be the result of the

experiment.

Problems 107

(a) Show that the result is equally likely to be

either heads or tails.

(b) Could we use a simpler procedure that contin-

ues to ﬂip the coin until the last two ﬂips are

different and then lets the result be the out-

come of the ﬁnal ﬂip?

3.59. Independent ﬂips of a coin that lands on heads

with probability p are made. What is the proba-

bility that the ﬁrst four outcomes are

(a) H, H, H, H?

(b) T, H, H, H?

H, H occurs before the pattern H, H, H, H?

Hint for part (c): How can the pattern H, H, H, H

occur ﬁrst?

3.60. The color of a person’s eyes is determined by a sin-

gle pair of genes. If they are both blue-eyed genes,

then the person will have blue eyes; if they are

both brown-eyed genes, then the person will have

brown eyes; and if one of them is a blue-eyed gene

and the other a brown-eyed gene, then the per-

son will have brown eyes. (Because of the latter

fact, we say that the brown-eyed gene is dominant

over the blue-eyed one.) A newborn child inde-

pendently receives one eye gene from each of its

parents, and the gene it receives from a parent is

equally likely to be either of the two eye genes of

that parent. Suppose that Smith and both of his

parents have brown eyes, but Smith’s sister has

blue eyes.

(a) What is the probability that Smith possesses a

blue-eyed gene?

(b) Suppose that Smith’s wife has blue eyes. What

is the probability that their ﬁrst child will have

blue eyes?

probability that their next child will also have

brown eyes?

3.61. Genes relating to albinism are denoted by A and

a. Only those people who receive the a gene from

both parents will be albino. Persons having the

gene pair A,

a are normal in appearance and,

because they can pass on the trait to their off-

spring, are called carriers. Suppose that a normal

couple has two children, exactly one of whom is

an albino. Suppose that the nonalbino child mates

with a person who is known to be a carrier for

albinism.

(a) What is the probability that their ﬁrst off-

spring is an albino?

(b) What is the conditional probability that their

second offspring is an albino given that their

ﬁrstborn is not?

3.62. Barbara and Dianne go target shooting. Suppose

that each of Barbara’s shots hits a wooden duck

target with probability p

, while each shot of

Dianne’s hits it with probability p

. Suppose that

they shoot simultaneously at the same target. If

the wooden duck is knocked over (indicating that

it was hit), what is the probability that

(a) both shots hit the duck?

(b) Barbara’s shot hit the duck?

What independence assumptions have you made?

3.63. A and B are involved in a duel. The rules of the

duel are that they are to pick up their guns and

shoot at each other simultaneously. If one or both

are hit, then the duel is over. If both shots miss,

then they repeat the process. Suppose that the

results of the shots are independent and that each

shot of A will hit B with probability p

, and each

shot of B will hit A with probability p

. What is

(a) the probability that A is not hit?

(b) the probability that both duelists are hit?

round of shots?

(d) the conditional probability that the duel ends

after the nth round of shots given that A is

not hit?

(e) the conditional probability that the duel ends

after the nth round of shots given that both

duelists are hit?

3.64. A true–false question is to be posed to a husband-

and-wife team on a quiz show. Both the husband

and the wife will independently give the correct

answer with probability p. Which of the following

is a better strategy for the couple?

(a) Choose one of them and let that person

answer the question.

(b) Have them both consider the question, and

then either give the common answer if they

agree or, if they disagree, ﬂip a coin to deter-

mine which answer to give.

3.65. In Problem 3.5, if p = .6 and the couple uses the

strategy in part (b), what is the conditional prob-

ability that the couple gives the correct answer

given that the husband and wife (a) agree? (b) dis-

agree?

3.66. The probability of the closing of the ith relay in the

circuits shown in Figure 3.4 is given by p

, i = 1, 2,

3, 4, 5. If all relays function independently, what is

the probability that a current ﬂows between A and

B for the respective circuits?

Hint for (b): Condition on whether relay 3 closes.

3.67. An engineering system consisting of n compo-

nents is said to be a k-out-of-n system (k … n)

if the system functions if and only if at least

k of the n components function. Suppose that

all components function independently of each

other.

(a) If the ith component functions with probabil-

ity P

, i = 1, 2, 3, 4, compute the probability

that a 2-out-of-4 system functions.

108 Chapter 3 Conditional Probability and Independence

A B

(a)

(b)

FIGURE 3.4: Circuits for Problem 3.66

(b) Repeat part (a) for a 3-out-of-5 system.

equal p (that is, P

= p, i = 1, 2, ..., n).

3.68. In Problem 3.65a, ﬁnd the conditional probability

that relays 1 and 2 are both closed given that a cur-

rent ﬂows from A to B.

3.69. A certain organism possesses a pair of each of 5

different genes (which we will designate by the

ﬁrst 5 letters of the English alphabet). Each gene

appears in 2 forms (which we designate by low-

ercase and capital letters). The capital letter will

be assumed to be the dominant gene, in the sense

that if an organism possesses the gene pair xX,

then it will outwardly have the appearance of the

X gene. For instance, if X stands for brown eyes

and x for blue eyes, then an individual having

either gene pair XX or xX will have brown eyes,

whereas one having gene pair xx will have blue

eyes. The characteristic appearance of an organ-

ism is called its phenotype, whereas its genetic

constitution is called its genotype. (Thus, 2 organ-

isms with respective genotypes aA, bB, cc, dD,

ee and AA, BB, cc, DD, ee would have different

genotypes but the same phenotype.) In a mating

between 2 organisms, each one contributes, at ran-

dom, one of its gene pairs of each type. The 5

contributions of an organism (one of each of the

5 types) are assumed to be independent and are

also independent of the contributions of the organ-

ism’s mate. In a mating between organisms hav-

ing genotypes aA, bB, cC, dD, eE and aa, bB, cc,

Dd, ee what is the probability that the progeny

will (i) phenotypically and (ii) genotypically

resemble

(a) the ﬁrst parent?

(b) the second parent?

(d) neither parent?

3.70. There is a 50–50 chance that the queen carries the

gene for hemophilia. If she is a carrier, then each

prince has a 50–50 chance of having hemophilia. If

the queen has had three princes without the dis-

ease, what is the probability that the queen is a

carrier? If there is a fourth prince, what is the prob-

ability that he will have hemophilia?

3.71. On the morning of September 30, 1982, the won–

lost records of the three leading baseball teams in

the Western Division of the National League were

as follows:

Team Won Lost

Atlanta Braves 87 72

San Francisco Giants 86 73

Los Angeles Dodgers 86 73

Each team had 3 games remaining. All 3 of the

Giants’ games were with the Dodgers, and the 3

remaining games of the Braves were against the

San Diego Padres. Suppose that the outcomes of

all remaining games are independent and each

game is equally likely to be won by either partic-

ipant. For each team, what is the probability that it

will win the division title? If two teams tie for ﬁrst

place, they have a playoff game, which each team

has an equal chance of winning.

3.72. A town council of 7 members contains a steering

committee of size 3. New ideas for legislation go

ﬁrst to the steering committee and then on to the

council as a whole if at least 2 of the 3 commit-

tee members approve the legislation. Once at the

full council, the legislation requires a majority vote

Problems 109

(of at least 4) to pass. Consider a new piece of

legislation, and suppose that each town council

member will approve it, independently, with prob-

ability p. What is the probability that a given steer-

ing committee member’s vote is decisive in the

sense that if that person’s vote were reversed,

then the ﬁnal fate of the legislation would be

reversed? What is the corresponding probability

for a given council member not on the steering

committee?

3.73. Suppose that each child born to a couple is equally

likely to be a boy or a girl, independently of the

sex distribution of the other children in the fam-

ily. For a couple having 5 children, compute the

probabilities of the following events:

(a) All children are of the same sex.

(b) The 3 eldest are boys and the others girls.

(d) The 2 oldest are girls.

(e) There is at least 1 girl.

3.74. A and B alternate rolling a pair of dice, stopping

either when A rolls the sum 9 or when B rolls the

sum 6. Assuming that A rolls ﬁrst, ﬁnd the proba-

bility that the ﬁnal roll is made by A.

3.75. In a certain village, it is traditional for the eldest

son (or the older son in a two-son family) and

his wife to be responsible for taking care of his

parents as they age. In recent years, however, the

women of this village, not wanting that responsi-

bility, have not looked favorably upon marrying an

eldest son.

(a) If every family in the village has two children,

what proportion of all sons are older sons?

(b) If every family in the village has three chil-

dren, what proportion of all sons are eldest

sons?

Assume that each child is, independently, equally

likely to be either a boy or a girl.

3.76. Suppose that E and F are mutually exclusive

events of an experiment. Show that if independent

trials of this experiment are performed, then E

will occur before F with probability P(E)/[P(E ) +

P(F)].

3.77. Consider an unending sequence of independent

trials, where each trial is equally likely to result in

any of the outcomes 1, 2, or 3. Given that outcome

3 is the last of the three outcomes to occur, ﬁnd the

conditional probability that

(a) the ﬁrst trial results in outcome 1;

(b) the ﬁrst two trials both result in outcome 1.

3.78. A and B play a series of games. Each game is inde-

pendently won by A with probability p and by B

with probability 1 − p. They stop when the total

number of wins of one of the players is two greater

than that of the other player. The player with the

greater number of total wins is declared the winner

of the series.

(a) Find the probability that a total of 4 games are

played.

(b) Find the probability that A is the winner of

the series.

3.79. In successive rolls of a pair of fair dice, what is the

probability of getting 2 sevens before 6 even num-

bers?

3.80. In a certain contest, the players are of equal skill

and the probability is

that a speciﬁed one of

the two contestants will be the victor. In a group

of 2

players, the players are paired off against

each other at random. The 2

n−1

winners are again

paired off randomly, and so on, until a single win-

ner remains. Consider two speciﬁed contestants, A

and B, and deﬁne the events A

, i … n, E by

: A plays in exactly i contests:

E: A and B never play each other.

(a) Find P(A

), i = 1, ..., n.

(b) Find P(E).

= P(E). Show that

− 1

− 2

− 1





n−1

and use this formula to check the answer you

obtained in part (b).

Hint:FindP(E) by conditioning on which of

the events A

, i = 1, ..., n occur. In simplifying

your answer, use the algebraic identity

n−1



i=1

i−1

1 − nx

n−1

+ (n − 1)x

(1 − x)

For another approach to solving this problem,

note that there are a total of 2

− 1 games

played.

(d) Explain why 2

− 1 games are played.

Number these games, and let B

denote the

event that A and B play each other in game

i, i = 1, ...,2

− 1.

(e) What is P(B

(f) Use part (e) to ﬁnd P (E ).

3.81. An investor owns shares in a stock whose present

value is 25. She has decided that she must sell her

stock if it goes either down to 10 or up to 40. If each

change of price is either up 1 point with probabil-

ity .55 or down 1 point with probability .45, and

the successive changes are independent, what is

the probability that the investor retires a winner?

3.82. A and B ﬂip coins. A starts and continues ﬂipping

until a tail occurs, at which point B starts ﬂipping

and continues until there is a tail. Then A takes

110 Chapter 3 Conditional Probability and Independence

over, and so on. Let P

be the probability of the

coin’s landing on heads when A ﬂips and P

when

B ﬂips. The winner of the game is the ﬁrst one

to get

(a) 2 heads in a row;

(b) a total of 2 heads;

(d) a total of 3 heads.

In each case, ﬁnd the probability that A wins.

3.83. Die A has 4 red and 2 white faces, whereas die B

has 2 red and 4 white faces. A fair coin is ﬂipped

once. If it lands on heads, the game continues with

die A; if it lands on tails, then die B is to be used.

(a) Show that the probability of red at any throw

(b) If the ﬁrst two throws result in red, what is the

probability of red at the third throw?

is the probability that it is die A that is being

used?

3.84. An urn contains 12 balls, of which 4 are white.

Three players—A, B,andC—successively draw

from the urn, A ﬁrst, then B,thenC,thenA,andso

on. The winner is the ﬁrst one to draw a white ball.

Find the probability of winning for each player if

(a) each ball is replaced after it is drawn;

(b) the balls that are withdrawn are not replaced.

3.85. Repeat Problem 3.84 when each of the 3 players

selects from his own urn. That is, suppose that

there are 3 different urns of 12 balls with 4 white

balls in each urn.

3.86. Let S ={1, 2, ..., n} and suppose that A and B are,

independently, equally likely to be any of the 2

subsets (including the null set and S itself) of S.

(a) Show that

P{A ( B}=





Hint:LetN(B) denote the number of ele-

ments in B.Use

P{A ( B}=



i=0

P{A ( B|N(B) = i}P{N(B) = i}

Show that P{AB = Ø}=





3.87. In Example 5e, what is the conditional probability

that the ith coin was selected given that the ﬁrst n

trials all result in heads?

3.88. In Laplace’s rule of succession (Example 5e), are

the outcomes of the successive ﬂips independent?

Explain.

3.89. A person tried by a 3-judge panel is declared guilty

if at least 2 judges cast votes of guilty. Suppose

that when the defendant is in fact guilty, each

judge will independently vote guilty with proba-

bility .7, whereas when the defendant is in fact

innocent, this probability drops to .2. If 70 per-

cent of defendants are guilty, compute the condi-

tional probability that judge number 3 votes guilty

given that

(a) judges 1 and 2 vote guilty;

(b) judges 1 and 2 cast 1 guilty and 1 not

guilty vote;

Let E

, i = 1, 2, 3 denote the event that judge

i casts a guilty vote. Are these events inde-

pendent. Are they conditionally independent?

Explain.

3.90. Suppose that n independent trials, each of which

results in any of the outcomes 0, 1, or 2, with

respective probabilities p

, p

,andp



i=0

= 1,

are performed. Find the probability that outcomes

1 and 2 both occur at least once.

THEORETICAL EXERCISES

3.1. Show that if P(A)>0, then

P(AB|A) Ú P(AB|A ∪ B)

3.2. Let A ( B. Express the following probabilities as

simply as possible:

P(A|B), P(A|B

), P(B|A), P(B|A

)

3.3. Consider a school community of m families, with n

of them having i children, i = 1, ..., k,



i=1

= m.

Consider the following two methods for choosing

a child:

1. Choose one of the m families at random and

then randomly choose a child from that family.

2. Choose one of the



i=1

children at random.

Show that method 1 is more likely than method 2

to result in the choice of a ﬁrstborn child.

Hint: In solving this problem, you will need to

show that



i=1



j=1



i=1



j=1

Theoretical Exercises 111

To do so, multiply the sums and show that, for all

pairs i, j, the coefﬁcient of the term n

is greater

in the expression on the left than in the one on the

right.

3.4. A ball is in any one of n boxes and is in the ith box

with probability P

. If the ball is in box i, a search of

that box will uncover it with probability α

. Show

that the conditional probability that the ball is in

box j, given that a search of box i did not uncover

it, is

1 − α

if j Z i

(1 − α

1 − α

if j = i

3.5. An event F is said to carry negative information

about an event E, and we write F R E,if

P(E|F) … P(E)

Prove or give counterexamples to the following

assertions:

(a) If F R E,thenE R F.

(b) If F R E and E R G,thenF R G.

Repeat parts (a), (b), and (c) when R is replaced

by Q, where we say that F carries positive informa-

tion about E, written F Q E,whenP(E|F) Ú P(E).

3.6. Prove that if E

, E

, ..., E

are independent

events, then

P(E

∪ E

∪ ··· ∪ E

) = 1 −



i=1

[1 − P(E

)]

3.7. (a) An urn contains n white and m black balls.

The balls are withdrawn one at a time until

only those of the same color are left. Show

that, with probability n/(n + m),theyareall

white.

Hint: Imagine that the experiment continues

until all the balls are removed, and consider

the last ball withdrawn.

(b) A pond contains 3 distinct species of ﬁsh,

which we will call the Red, Blue, and Green

ﬁsh. There are r Red, b Blue, and g Green ﬁsh.

Suppose that the ﬁsh are removed from the

pond in a random order. (That is, each selec-

tion is equally likely to be any of the remain-

ing ﬁsh.) What is the probability that the Red

ﬁsh are the ﬁrst species to become extinct in

the pond?

Hint:WriteP{R}=P{RBG}+P{RGB},

and compute the probabilities on the right

by ﬁrst conditioning on the last species to be

removed.

3.8. Let A, B,andC be events relating to the experi-

ment of rolling a pair of dice.

(a) If

P(A|C)>P(B|C) and P(A|C

)>P(B|C

)

either prove that P(A)>P(B) or give a coun-

terexample by deﬁning events A, B,andC for

which that relationship is not true.

(b) If

P(A|C)>P(A|C

) and P(B|C)>P(B|C

)

either prove that P(AB|C)>P(AB|C

) or

give a counterexample by deﬁning events

A, B,andC for which that relationship is not

true.

Hint:LetC be the event that the sum of a pair of

dice is 10; let A be the event that the ﬁrst die lands

on 6; let B be the event that the second die lands

on 6.

3.9. Consider two independent tosses of a fair coin. Let

A be the event that the ﬁrst toss results in heads, let

B be the event that the second toss results in heads,

and let C be the event that in both tosses the coin

lands on the same side. Show that the events A, B,

and C are pairwise independent—that is, A and B

are independent, A and C are independent, and B

and C are independent—but not independent.

3.10. Consider a collection of n individuals. Assume that

each person’s birthday is equally likely to be any of

the 365 days of the year and also that the birthdays

are independent. Let A

i,j

, i Z j, denote the event

that persons i and j have the same birthday. Show

that these events are pairwise independent, but not

independent. That is, show that A

i,j

and A

r,s

are

independent, but the





events A

i,j

, i Z j are not

independent.

3.11. In each of n independent tosses of a coin, the coin

lands on heads with probability p. How large need

n be so that the probability of obtaining at least

one head is at least

3.12. Show that 0 … a

… 1, i = 1, 2, ...,then



i=1

⎡

⎢

⎣

i−1



j=1

(1 − a

)

⎤

⎥

⎦



i=1

(1 − a

) = 1

Hint: Suppose that an inﬁnite number of coins are

to be ﬂipped. Let a

be the probability that the ith

coin lands on heads, and consider when the ﬁrst

head occurs.

3.13. The probability of getting a head on a single toss

of a coin is p. Suppose that A starts and continues

to ﬂip the coin until a tail shows up, at which point

112 Chapter 3 Conditional Probability and Independence

B starts ﬂipping. Then B continues to ﬂip until a

tail comes up, at which point A takes over, and so

on. Let P

n,m

denote the probability that A accu-

mulates a total of n heads before B accumulates

m. Show that

n,m

= pP

n−1,m

+ (1 − p)(1 − P

m,n

)

∗

3.14. Suppose that you are gambling against an inﬁnitely

rich adversary and at each stage you either win

or lose 1 unit with respective probabilities p and

1 − p. Show that the probability that you eventu-

ally go broke is

1ifp …

(q/p)

if p >

where q = 1 − p and where i is your initial fortune.

3.15. Independent trials that result in a success with

probability p are successively performed until a

total of r successes is obtained. Show that the prob-

ability that exactly n trials are required is



n − 1

r − 1



(1 − p)

n−r

Use this result to solve the problem of the points

(Example 4j).

Hint:Inorderforittotaken trials to obtain r suc-

cesses, how many successes must occur in the ﬁrst

n − 1 trials?

3.16. Independent trials that result in a success with

probability p and a failure with probability 1 −

p are called Bernoulli trials.LetP

denote the

probability that n Bernoulli trials result in an even

number of successes (0 being considered an even

number). Show that

= p(1 − P

n−1

) + (1 − p)P

n−1

n Ú 1

and use this formula to prove (by induction) that

1 + (1 − 2p)

3.17. Suppose that n independent trials are performed,

with trial i being a success with probability 1/(2i +

1).LetP

denote the probability that the total

number of successes that result is an odd number.

(a) Find P

for n = 1, 2, 3, 4, 5.

(b) Conjecture a general formula for P

in terms of P

n−1

(d) Verify that your conjecture in part (b) satisﬁes

the recursive formula in part (d). Because the

recursive formula has a unique solution, this

then proves that your conjecture is correct.

3.18. Let Q

denote the probability that no run of 3 con-

secutive heads appears in n tosses of a fair coin.

Show that

n−1

n−2

n−3

= Q

= 1

Find Q

Hint: Condition on the ﬁrst tail.

3.19. Consider the gambler’s ruin problem, with the

exception that A and B agree to play no more than

n games. Let P

n,i

denote the probability that A

winds up with all the money when A starts with

i and B starts with N − i. Derive an equation

for P

n,i

in terms of P

n−1, i+1

and P

n−1, i−1

,and

compute P

7, 3

, N = 5.

3.20. Consider two urns, each containing both white

and black balls. The probabilities of drawing white

balls from the ﬁrst and second urns are, respec-

tively, p and p



. Balls are sequentially selected with

replacement as follows: With probability α,aball

is initially chosen from the ﬁrst urn, and with prob-

ability 1 − α, it is chosen from the second urn. The

subsequent selections are then made according to

the rule that whenever a white ball is drawn (and

replaced), the next ball is drawn from the same

urn, but when a black ball is drawn, the next ball is

taken from the other urn. Let α

denote the prob-

ability that the nth ball is chosen from the ﬁrst urn.

Show that

n+1

= α

(p + p



− 1) + 1 − p



n Ú 1

and use this formula to prove that

1 − p



2 − p − p





α −

1 − p



2 − p − p





* (p + p



− 1)

n−1

Let P

denote the probability that the nth

ball selected is white. Find P

. Also, compute

lim

n→q

and lim

n→q

3.21. The Ballot Problem. In an election, candidate A

receives n votes and candidate B receives m votes,

where n > m. Assuming that all of the (n +

m)!/n! m! orderings of the votes are equally likely,

let P

n,m

denote the probability that A is always

ahead in the counting of the votes.

(a) Compute P

2,1

, P

3,1

, P

3,2

, P

4,1

, P

4,2

, P

4,3

(b) Find P

n,1

, P

n,2

(b), conjecture the value of P

n,m

(d) Derive a recursion for P

n,m

in terms of P

n−1,m

and P

n,m−1

by conditioning on who receives

the last vote.

Theoretical Exercises 113

(e) Use part (d) to verify your conjecture in part

3.22. As a simpliﬁed model for weather forecasting, sup-

pose that the weather (either wet or dry) tomor-

row will be the same as the weather today with

probability p. Show that the weather is dry on Jan-

uary 1, then P

, the probability that it will be dry n

days later, satisﬁes

= (2p − 1)P

n−1

+ (1 − p) n Ú 1

= 1

Prove that

(2p − 1)

n Ú 0

3.23. A bag contains a white and b black balls. Balls

are chosen from the bag according to the following

method:

1. A ball is chosen at random and is discarded.

2. A second ball is then chosen. If its color is

different from that of the preceding ball, it is

replaced in the bag and the process is repeated

from the beginning. If its color is the same, it is

discarded and we start from step 2.

In other words, balls are sampled and discarded

until a change in color occurs, at which point the

last ball is returned to the urn and the process

starts anew. Let P

a,b

denote the probability that

the last ball in the bag is white. Prove that

a,b

Hint: Use induction on k K a + b.

∗

3.24. A round-robin tournament of n contestants is a

tournament in which each of the





pairs of

contestants play each other exactly once, with the

outcome of any play being that one of the contes-

tants wins and the other loses. For a ﬁxed integer

k, k < n, a question of interest is whether it is pos-

sible that the tournament outcome is such that, for

every set of k players, there is a player who beat

each member of that set. Show that if







1 −







n−k

< 1

then such an outcome is possible.

Hint: Suppose that the results of the games are

independent and that each game is equally likely

to be won by either contestant. Number the





sets of k contestants, and let B

denote the event

that no contestant beat all of the k players in

the ith set. Then use Boole’s inequality to bound







3.25. Prove directly that

P(E|F) = P(E|FG)P(G|F) + P(E|FG

)P(G

|F)

3.26. Prove the equivalence of Equations (5.11) and

(5.12).

3.27. Extend the deﬁnition of conditional independence

to more than 2 events.

3.28. Prove or give a counterexample. If E

and E

are

independent, then they are conditionally indepen-

dent given F.

3.29. In Laplace’s rule of succession (Example 5e),

show that if the ﬁrst n ﬂips all result in heads,

then the conditional probability that the next m

ﬂips also result in all heads is (n + 1)/(n +

m + 1).

3.30. In Laplace’s rule of succession (Example 5e), sup-

pose that the ﬁrst n ﬂips resulted in r heads and

n − r tails. Show that the probability that the

(n + 1)st ﬂip turns up heads is (r + 1)/(n + 2).To

do so, you will have to prove and use the identity

(1 − y)

dy =

n!m!

(n + m + 1)!

Hint: To prove the identity, let C(n, m) =

(1 − y)

dy. Integrating by parts yields

C(n, m) =

n + 1

C(n + 1, m − 1)

Starting with C(n,0) = 1/(n + 1), prove the iden-

tity by induction on m.

3.31. Suppose that a nonmathematical, but philosophi-

cally minded, friend of yours claims that Laplace’s

rule of succession must be incorrect because it can

lead to ridiculous conclusions. “For instance,” says

he, “the rule states that if a boy is 10 years old,

having lived 10 years, the boy has probability

living another year. On the other hand, if the boy

has an 80-year-old grandfather, then, by Laplace’s

rule, the grandfather has probability

of sur-

viving another year. However, this is ridiculous.

Clearly, the boy is more likely to survive an addi-

tional year than the grandfather is.” How would

you answer your friend?

114 Chapter 3 Conditional Probability and Independence

SELF-TEST PROBLEMS AND EXERCISES

3.1. In a game of bridge, West has no aces. What is the

probability of his partner’s having (a) no aces? (b)

2 or more aces? (c) What would the probabilities

be if West had exactly 1 ace?

3.2. The probability that a new car battery functions

for over 10,000 miles is .8, the probability that it

functions for over 20,000 miles is .4, and the prob-

ability that it functions for over 30,000 miles is .1. If

a new car battery is still working after 10,000 miles,

what is the probability that

(a) its total life will exceed 20,000 miles?

(b) its additional life will exceed 20,000 miles?

3.3. How can 20 balls, 10 white and 10 black, be put

into two urns so as to maximize the probability of

drawing a white ball if an urn is selected at random

and a ball is drawn at random from it?

3.4. Urn A contains 2 white balls and 1 black ball,

whereas urn B contains 1 white ball and 5 black

balls. A ball is drawn at random from urn A and

placed in urn B. A ball is then drawn from urn B.

It happens to be white. What is the probability that

the ball transferred was white?

3.5. An urn has r red and w white balls that are ran-

domly removed one at a time. Let R

be the event

that the ith ball removed is red. Find

(a) P(R

)

(b) P(R

)

3.6. An urn contains b black balls and r red balls. One

of the balls is drawn at random, but when it is

put back in the urn, c additional balls of the same

color are put in with it. Now, suppose that we

draw another ball. Show that the probability that

the ﬁrst ball was black, given that the second ball

drawn was red, is b/(b + r + c).

3.7. A friend randomly chooses two cards, without

replacement, from an ordinary deck of 52 playing

cards. In each of the following situations, deter-

mine the conditional probability that both cards

are aces.

(a) You ask your friend if one of the cards is the

ace of spades, and your friend answers in the

afﬁrmative.

(b) You ask your friend if the ﬁrst card selected

is an ace, and your friend answers in the afﬁr-

mative.

selected is an ace, and your friend answers in

the afﬁrmative.

(d) You ask your friend if either of the cards

selected is an ace, and your friend answers in

the afﬁrmative.

3.8. Show that

P(H|E)

P(G|E)

P(H)

P(G)

P(E|H)

P(E|G)

Suppose that, before new evidence is observed, the

hypothesis H is three times as likely to be true as

is the hypothesis G. If the new evidence is twice

as likely when G is true than it is when H is true,

which hypothesis is more likely after the evidence

has been observed?

3.9. You ask your neighbor to water a sickly plant

while you are on vacation. Without water, it will

die with probability .8; with water, it will die with

probability .15. You are 90 percent certain that

your neighbor will remember to water the plant.

(a) What is the probability that the plant will be

alive when you return?

(b) If the plant is dead upon your return, what is

the probability that your neighbor forgot to

water it?

3.10. Six balls are to be randomly chosen from an urn

containing 8 red, 10 green, and 12 blue balls.

(a) What is the probability at least one red ball is

chosen?

(b) Given that no red balls are chosen, what is the

conditional probability that there are exactly

2 green balls among the 6 chosen?

3.11. A type C battery is in working condition with prob-

ability .7, whereas a type D battery is in work-

ing condition with probability .4. A battery is ran-

domly chosen from a bin consisting of 8 type C and

6 type D batteries.

(a) What is the probability that the battery

works?

(b) Given that the battery does not work, what is

the conditional probability that it was a type

C battery?

3.12. Maria will take two books with her on a trip. Sup-

pose that the probability that she will like book 1

is .6, the probability that she will like book 2 is .5,

and the probability that she will like both books

is .4. Find the conditional probability that she will

like book 2 given that she did not like book 1.

3.13. Balls are randomly removed from an urn that ini-

tially contains 20 red and 10 blue balls.

(a) What is the probability that all of the red balls

are removed before all of the blue ones have

been removed?

Now suppose that the urn initially contains 20

red, 10 blue, and 8 green balls.

(b) Now what is the probability that all of the red

balls are removed before all of the blue ones

have been removed?

Self-Test Problems and Exercises 115

depleted in the order blue, red, green?

(d) What is the probability that the group of blue

balls is the ﬁrst of the three groups to be

removed?

3.14. A coin having probability .8 of landing on heads

is ﬂipped. A observes the result—either heads or

tails—and rushes off to tell B. However, with prob-

ability .4, A will have forgotten the result by the

time he reaches B.IfA has forgotten, then, rather

than admitting this to B, he is equally likely to tell

B that the coin landed on heads or that it landed

tails. (If he does remember, then he tells B the cor-

rect result.)

(a) What is the probability that B is told that the

coin landed on heads?

(b) What is the probability that B is told the cor-

rect result?

heads, what is the probability that it did in fact

land on heads?

3.15. In a certain species of rats, black dominates over

brown. Suppose that a black rat with two black

parents has a brown sibling.

(a) What is the probability that this rat is a pure

black rat (as opposed to being a hybrid with

one black and one brown gene)?

(b) Suppose that when the black rat is mated with

a brown rat, all 5 of their offspring are black.

Now what is the probability that the rat is a

pure black rat?

3.16. (a) In Problem 3.65b, ﬁnd the probability that a

current ﬂows from A to B, by conditioning on

whether relay 1 closes.

(b) Find the conditional probability that relay 3 is

closed given that a current ﬂows from A to B.

3.17. For the k-out-of-n system described in

Problem 3.67, assume that each component

independently works with probability

.Findthe

conditional probability that component 1 is work-

ing, given that the system works, when

(a) k = 1, n = 2;

(b) k = 2, n = 3.

3.18. Mr. Jones has devised a gambling system for win-

ning at roulette. When he bets, he bets on red and

places a bet only when the 10 previous spins of

the roulette have landed on a black number. He

reasons that his chance of winning is quite large

because the probability of 11 consecutive spins

resulting in black is quite small. What do you think

of this system?

3.19. Three players simultaneously toss coins. The coin

tossed by A(B)[C] turns up heads with probability

)[P

]. If one person gets an outcome differ-

ent from those of the other two, then he is the odd

man out. If there is no odd man out, the players ﬂip

again and continue to do so until they get an odd

man out. What is the probability that A will be the

odd man?

3.20. Suppose that there are n possible outcomes of

a trial, with outcome i resulting with probability

, i = 1, ..., n,



i=1

= 1. If two independent tri-

als are observed, what is the probability that the

result of the second trial is larger than that of the

ﬁrst?

3.21. If A ﬂips n + 1andB ﬂips n fair coins, show that

the probability that A gets more heads than B is

Hint: Condition on which player has more heads

after each has ﬂipped n coins. (There are three

possibilities.)

3.22. Prove or give counterexamples to the following

statements:

(a) If E is independent of F and E is independent

of G,thenE is independent of F ∪ G.

(b) If E is independent of F,andE is independent

of G,andFG = Ø, then E is independent of

F ∪ G.

of G,andE is independent of FG,thenG is

independent of EF.

3.23. Let A and B be events having positive probabil-

ity. State whether each of the following statements

is (i) necessarily true, (ii) necessarily false, or (iii)

possibly true.

(a) If A and B are mutually exclusive, then they

are independent.

(b) If A and B are independent, then they are

mutually exclusive.

6, and A and B are mutually

exclusive.

(d) P(A) = P(B) = .6, and A and B are indepen-

dent.

3.24. Rank the following from most likely to least likely

to occur:

1. A fair coin lands on heads.

2. Three independent trials, each of which is a suc-

cess with probability .8, all result in successes.

3. Seven independent trials, each of which is a suc-

cess with probability .9, all result in successes.

3.25. Two local factories, A and B, produce radios. Each

radio produced at factory A is defective with prob-

ability .05, whereas each one produced at factory B

is defective with probability .01. Suppose you pur-

chase two radios that were produced at the same

factory, which is equally likely to have been either

factory A or factory B. If the ﬁrst radio that you

check is defective, what is the conditional proba-

bility that the other one is also defective?

116 Chapter 3 Conditional Probability and Independence

3.26. Show that if P(A|B) = 1, then P(B

) = 1.

3.27. An urn initially contains 1 red and 1 blue ball.

At each stage, a ball is randomly withdrawn and

replaced by two other balls of the same color.

(For instance, if the red ball is initially chosen,

thentherewouldbe2redand1blueballin

the urn when the next selection occurs.) Show

by mathematical induction that the probability

that there are exactly i red balls in the urn

after n stages have been completed is

n+1

,1 …

i … n + 1.

3.28. A total of 2n cards, of which 2 are aces, are

to be randomly divided among two players, with

each player receiving n cards. Each player is then

to declare, in sequence, whether he or she has

received any aces. What is the conditional proba-

bility that the second player has no aces, given that

the ﬁrst player declares in the afﬁrmative, when

(a) n = 2? (b) n = 10? (c) n = 100? To what

does the probability converge as n goes to inﬁnity?

Why?

3.29. There are n distinct types of coupons, and

each coupon obtained is, independently of prior

types collected, of type i with probability p



i=1

= 1.

(a) If n coupons are collected, what is the proba-

bility that one of each type is obtained?

(b) Now suppose that p

= p

=···=p

= 1/n.

Let E

be the event that there are no type

i coupons among the n collected. Apply the

inclusion–exclusion identity for the probabil-

ity of the union of events to P(∪

) to prove

the identity

n! =



k=0

(−1)





(n − k)

3.30. Show that, for any events E and F,

P(E|E ∪ F) Ú P(E|F)

Hint: Compute P(E|E ∪ F) by conditioning on

whether F occurs.

CHAPTER 4

Random Variables

4.1 RANDOM VARIABLES

4.2 DISCRETE RANDOM VARIABLES

4.3 EXPECTED VALUE

4.4 EXPECTATION OF A FUNCTION OF A RANDOM VARIABLE

4.5 VARIANCE

4.6 THE BERNOULLI AND BINOMIAL RANDOM VARIABLES

4.7 THE POISSON RANDOM VARIABLE

4.8 OTHER DISCRETE PROBABILITY DISTRIBUTIONS

4.9 EXPECTED VALUE OF SUMS OF RANDOM VARIABLES

4.10 PROPERTIES OF THE CUMULATIVE DISTRIBUTION FUNCTION

4.1 RANDOM VARIABLES

Frequently, when an experiment is performed, we are interested mainly in some func-

tion of the outcome as opposed to the actual outcome itself. For instance, in tossing

dice, we are often interested in the sum of the two dice and are not really concerned

about the separate values of each die. That is, we may be interested in knowing

that the sum is 7 and may not be concerned over whether the actual outcome was

(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), or (6, 1). Also, in ﬂipping a coin, we may be inter-

ested in the total number of heads that occur and not care at all about the actual

head–tail sequence that results. These quantities of interest, or, more formally, these

real-valued functions deﬁned on the sample space, are known as random variables.

Because the value of a random variable is determined by the outcome of the exper-

iment, we may assign probabilities to the possible values of the random variable.

EXAMPLE 1a

Suppose that our experiment consists of tossing 3 fair coins. If we let Y denote the

number of heads that appear, then Y is a random variable taking on one of the values

0, 1, 2, and 3 with respective probabilities

P{Y = 0}=P{(T, T, T)}=

P{Y = 1}=P{(T, T, H), (T, H, T), (H, T, T)}=

P{Y = 2}=P{(T, H, H), (H, T, H), (H, H, T)}=

P{Y = 3}=P{(H, H, H)}=

117

118 Chapter 4 Random Variables

Since Y must take on one of the values 0 through 3, we must have

1 = P

⎛

⎝



i=0

{Y = i}

⎞

⎠



i=0

P{Y = i}

which, of course, is in accord with the preceding probabilities. .

EXAMPLE 1b

Three balls are to be randomly selected without replacement from an urn contain-

ing 20 balls numbered 1 through 20. If we bet that at least one of the balls that are

drawn has a number as large as or larger than 17, what is the probability that we

win the bet?

Solution. Let X denote the largest number selected. Then X is a random variable

taking on one of the values 3, 4, ..., 20. Furthermore, if we suppose that each of the





possible selections are equally likely to occur, then

P{X = i}=



i − 1







i = 3, ..., 20 (1.1)

Equation (1.1) follows because the number of selections that result in the event

{X = i} is just the number of selections that result in the ball numbered i and two

of the balls numbered 1 through i − 1 being chosen. Because there are clearly







i − 1



such selections, we obtain the probabilities expressed in Equation (1.1),

from which we see that

P{X = 20}=









= .150

P{X = 19}=









380

L .134

P{X = 18}=









285

L .119

P{X = 17}=









L .105

Section 4.1 Random Variables 119

Hence, since the event {X Ú 17} is the union of the disjoint events {X = i},

i = 17, 18, 19, 20, it follows that the probability of our winning the bet is given by

P{X Ú 17} L .105 + .119 + .134 + .150 = .508

EXAMPLE 1c

Independent trials consisting of the ﬂipping of a coin having probability p of coming

up heads are continually performed until either a head occurs or a total of n ﬂips is

made. If we let X denote the number of times the coin is ﬂipped, then X is a random

variable taking on one of the values 1, 2, 3, ..., n with respective probabilities

P{X = 1}=P{H}=p

P{X = 2}=P{(T, H)}=(1 − p)p

P{X = 3}=P{(T, T, H)

}=(1 − p)

P{X = n − 1}=P{(T, T, ..., T,



 

n−2

H)}=(1 − p)

n−2

P{X = n}=P{(T, T, ..., T,



 

n−1

T), (T, T, ..., T,



 

n−1

H)}=(1 − p)

n−1

As a check, note that

⎛

⎝



i=1

{X = i}

⎞

⎠



i=1

P{X = i}

n−1



i=1

p(1 − p)

i−1

+ (1 − p)

n−1

= p



1 − (1 − p)

n−1

1 − (1 − p)



+ (1 − p)

n−1

= 1 − (1 − p)

n−1

+ (1 − p)

n−1

= 1

EXAMPLE 1d

Three balls are randomly chosen from an urn containing 3 white, 3 red, and 5 black

balls. Suppose that we win $1 for each white ball selected and lose $1 for each red ball

120 Chapter 4 Random Variables

selected. If we let X denote our total winnings from the experiment, then X is a ran-

dom variable taking on the possible values 0, ;1, ;2, ;3 with respective probabilities

P{X = 0}=















165

P{X = 1}=P{X =−1}=

















165

P{X = 2}=P{X =−2}=











165

P{X = 3}=P{X =−3}=









165

These probabilities are obtained, for instance, by noting that in order for X to

equal 0, either all 3 balls selected must be black or 1 ball of each color must be

selected. Similarly, the event {X = 1} occurs either if 1 white and 2 black balls are

selected or if 2 white and 1 red is selected. As a check, we note that



i=0

P{X = i}+



i=1

P{X =−i}=

55 + 39 + 15 + 1 + 39 + 15 + 1

165

= 1

The probability that we win money is given by



i=1

P{X = i}=

165

EXAMPLE 1e

Suppose that there are N distinct types of coupons and that each time one obtains a

coupon, it is, independently of previous selections, equally likely to be any one of the

N types. One random variable of interest is T, the number of coupons that needs to

be collected until one obtains a complete set of at least one of each type. Rather than

derive P{T = n} directly, let us start by considering the probability that T is greater

than n. To do so, ﬁx n and deﬁne the events A

, A

, ..., A

as follows: A

is the event

Section 4.1 Random Variables 121

that no type j coupon is contained among the ﬁrst n coupons collected, j = 1, ..., N.

Hence,

P{T > n}=P

⎛

⎜

⎝



j=1

⎞

⎟

⎠



P(A

) −



P(A

) + ···

+ (−1)

k+1



<···<j

P(A

···A

) ···

+ (−1)

N+1

P(A

···A

)

Now, A

will occur if each of the n coupons collected is not of type j. Since each of the

coupons will not be of type j with probability (N − 1)/N, we have, by the assumed

independence of the types of successive coupons,

P(A

) =



N − 1



Also, the event A

will occur if none of the ﬁrst n coupons collected is of either

type j

or type j

. Thus, again using independence, we see that

P(A

) =



N − 2



The same reasoning gives

P(A

···A

) =



N − k



and we see that, for n > 0,

P{T > n}=N



N − 1



−





N − 2







N − 3



− ···

+ (−1)



N − 1





N−1



i=1





N − i



(−1)

i+1

(1.2)

The probability that T equals n can now be obtained from the preceding formula by

the use of

P{T > n − 1}=P {T = n}+P{T > n}

or, equivalently,

P{T = n}=P {T > n − 1}−P{T > n}

Another random variable of interest is the number of distinct types of coupons

that are contained in the ﬁrst n selections—call this random variable D

. To compute

122 Chapter 4 Random Variables

P{D

= k}, let us start by ﬁxing attention on a particular set of k distinct types,

and let us then determine the probability that this set constitutes the set of distinct

types obtained in the ﬁrst n selections. Now, in order for this to be the situation, it is

necessary and sufﬁcient that, of the ﬁrst n coupons obtained,

A: each is one of these k types.

B: each of these k types is represented.

Now, each coupon selected will be one of the k types with probability k/N,sothe

probability that A will be valid is (k/N)

. Also, given that a coupon is of one of the k

types under consideration, it is easy to see that it is equally likely to be of any one of

these k types. Hence, the conditional probability of B given that A occurs is the same

as the probability that a set of n coupons, each equally likely to be any of k possible

types, contains a complete set of all k types. But this is just the probability that the

number needed to amass a complete set, when choosing among k types, is less than

or equal to n and is thus obtainable from Equation (1.2) with k replacing N. Thus,

we have

P(A) =





P(B|A) = 1 −

k−1



i=1





k − i



(−1)

i+1

Finally, as there are





possible choices for the set of k types, we arrive at

P{D

= k}=





P(AB)







⎡

⎣

1 −

k−1



i=1





k − i



(−1)

i+1

⎤

⎦

Remark. Since one must collect at least N coupons to obtain a compete set, it

follows that P{T > n}=1ifn < N. Therefore, from Equation (1.2), we obtain the

interesting combinatorial identity that, for integers 1 … n < N,

N−1



i=1





N − i



(−1)

i+1

= 1

which can be written as

N−1



i=0





N − i



(−1)

i+1

= 0

or, upon multiplying by (−1)

and letting j = N − i,



j=1





(−1)

j−1

= 01… n < N

Section 4.2 Discrete Random Variables 123

For a random variable X, the function F deﬁned by

F(x) = P{X … x}−q < x < q

is called the cumulative distribution function, or, more simply, the distribution func-

tion,ofX. Thus, the distribution function speciﬁes, for all real values x, the probability

that the random variable is less than or equal to x.

Now, suppose that a … b. Then, because the event {X … a} is contained in the

event {X … b}, it follows that F(a), the probability of the former, is less than or

equal to F(b), the probability of the latter. In other words, F(x) is a nondecreas-

ing function of x. Other general properties of the distribution function are given in

Section 4.10.

4.2 DISCRETE RANDOM VARIABLES

A random variable that can take on at most a countable number of possible values is

said to be discrete. For a discrete random variable X, we deﬁne the probability mass

function p(a) of X by

p(a) = P{X = a}

The probability mass function p(a)

is positive for at most a countable number of val-

ues of a. That is, if X must assume one of the values x

, x

, ...,then

p(x

) Ú 0fori = 1, 2, ...

p(x) = 0 for all other values of x

Since X must take on one of the values x

, we have



i=1

p(x

) = 1

It is often instructive to present the probability mass function in a graphical format

by plotting p(x

) on the y-axis against x

on the x-axis. For instance, if the probability

mass function of X is

p(0) =

p(1) =

p(2) =

–

(x)

–

FIGURE 4.1

124 Chapter 4 Random Variables

—

p(x)

—

3 4 567 8 9 10 11 12

FIGURE 4.2

we can represent this function graphically as shown in Figure 4.1. Similarly, a graph

of the probability mass function of the random variable representing the sum when

two dice are rolled looks like Figure 4.2.

EXAMPLE 2a

The probability mass function of a random variable X is given by p(i) = cλ

/i!,

i = 0, 1, 2, ..., where λ is some positive value. Find (a) P{X = 0} and (b) P{X > 2}.

Solution. Since



i=0

p(i) = 1, we have



i=0

= 1

which, because e



i=0

/i!, implies that

= 1orc = e

−λ

Hence,

(a) P{X = 0}=e

−λ

/0! = e

−λ

(b) P{X > 2}=1 − P{X … 2}=1 − P{X = 0}−P{X = 1}

− P{X = 2}

= 1 − e

−λ

− λe

−λ

−

−λ

The cumulative distribution function F can be expressed in terms of p(a) by

F(a) =



all x … a

p(x)

If X is a discrete random variable whose possible values are x

, x

, ..., where

< x

< ···, then the distribution function F of X is a step function. That is,

Section 4.3 Expected Value 125

the value of F is constant in the intervals [x

i−1

, x

) and then takes a step (or jump) of

size p(x

) at x

. For instance, if X has a probability mass function given by

p(1) =

p(2) =

p(3) =

p(4) =

then its cumulative distribution function is

F(a) =

⎧

⎪

⎨

⎪

⎩

0 a < 1

1 … a < 2

2 … a < 3

3 … a < 4

14… a

This function is depicted graphically in Figure 4.3.

–

123

(

)

–

FIGURE 4.3

Note that the size of the step at any of the values 1, 2, 3, and 4 is equal to the

probability that X assumes that particular value.

4.3 EXPECTED VALUE

One of the most important concepts in probability theory is that of the expectation

of a random variable. If X is a discrete random variable having a probability mass

function p(x), then the expectation,ortheexpected value,ofX, denoted by E[X], is

deﬁned by

E[X] =



x:p(x)>0

xp(x)

In words, the expected value of X is a weighted average of the possible values that

X can take on, each value being weighted by the probability that X assumes it. For

instance, on the one hand, if the probability mass function of X is given by

p(0) =

= p(1)

then

E[X] = 0





+ 1





126 Chapter 4 Random Variables

is just the ordinary average of the two possible values, 0 and 1, that X can assume.

On the other hand, if

p(0) =

p(1) =

then

E[X] = 0





+ 1





is a weighted average of the two possible values 0 and 1, where the value 1 is given

twice as much weight as the value 0, since p(1) = 2p(0).

Another motivation of the deﬁnition of expectation is provided by the frequency

interpretation of probabilities. This interpretation (partially justiﬁed by the strong

law of large numbers, to be presented in Chapter 8) assumes that if an inﬁnite sequence

of independent replications of an experiment is performed, then, for any event E,

the proportion of time that E occurs will be P(E). Now, consider a random vari-

able X that must take on one of the values x

, x

, ...x

with respective probabilities

p(x

), p(x

), ..., p(x

), and think of X as representing our winnings in a single game of

chance. That is, with probability p(x

) we shall win x

units i = 1, 2, ..., n. By the fre-

quency interpretation, if we play this game continually, then the proportion of time

that we win x

will be p(x

). Since this is true for all i, i = 1, 2, ..., n, it follows that our

average winnings per game will be



i=1

p(x

) = E[X]

EXAMPLE 3a

Find E[X], where X is the outcome when we roll a fair die.

Solution. Since p(1) = p(2) = p(3) = p(4) = p(5) = p(6) =

, we obtain

E[X] = 1





+ 2





+ 3





+ 4





+ 5





+ 6





EXAMPLE 3b

We say that I is an indicator variable for the event A if

I =

1ifA occurs

0ifA

occurs

Find E[I].

Solution. Since p(1) = P(A), p(0) = 1 − P(A), we have

E[I] = P(A)

That is, the expected value of the indicator variable for the event A is equal to the

probability that A occurs. .

Section 4.3 Expected Value 127

EXAMPLE 3c

A contestant on a quiz show is presented with two questions, questions 1 and 2, which

he is to attempt to answer in some order he chooses. If he decides to try question i

ﬁrst, then he will be allowed to go on to question j, j Z i, only if his answer to question

i is correct. If his initial answer is incorrect, he is not allowed to answer the other ques-

tion. The contestant is to receive V

dollars if he answers question i correctly, i = 1, 2.

For instance, he will receive V

+ V

dollars if he answers both questions correctly.

If the probability that he knows the answer to question i is P

, i = 1, 2, which question

should he attempt to answer ﬁrst so as to maximize his expected winnings? Assume

that the events E

, i = 1, 2, that he knows the answer to question i are independent

events.

Solution. On the one hand, if he attempts to answer question 1 ﬁrst, then he will win

0 with probability 1 − P

with probability P

(1 − P

)

+ V

with probability P

Hence, his expected winnings in this case will be

(1 − P

) + (V

+ V

On the other hand, if he attempts to answer question 2 ﬁrst, his expected winnings

will be

(1 − P

) + (V

+ V

Therefore, it is better to try question 1 ﬁrst if

(1 − P

) Ú V

(1 − P

)

or, equivalently, if

1 − P

For example, if he is 60 percent certain of answering question 1, worth $200, correctly

and he is 80 percent certain of answering question 2, worth $100, correctly, then he

should attempt to answer question 2 ﬁrst because

400 =

(100)(.8)

(200)(.6)

= 300

EXAMPLE 3d

A school class of 120 students is driven in 3 buses to a symphonic performance. There

are 36 students in one of the buses, 40 in another, and 44 in the third bus. When the

buses arrive, one of the 120 students is randomly chosen. Let X denote the number

of students on the bus of that randomly chosen student, and ﬁnd E[X].

Solution. Since the randomly chosen student is equally likely to be any of the 120

students, it follows that

P{X = 36}=

120

P{X = 40}=

120

P{X = 44}=

120

Hence,

E[X] = 36





+ 40





+ 44





1208

= 40.2667

128 Chapter 4 Random Variables

However, the average number of students on a bus is 120/3 = 40, showing that the

expected number of students on the bus of a randomly chosen student is larger than

the average number of students on a bus. This is a general phenomenon, and it occurs

because the more students there are on a bus, the more likely it is that a randomly

chosen student would have been on that bus. As a result, buses with many students

are given more weight than those with fewer students. (See Self-Test Problem 4.) .

Remark. The probability concept of expectation is analogous to the physical con-

ceptofthecenter of gravity of a distribution of mass. Consider a discrete random

variable X having probability mass function p(x

), i Ú 1. If we now imagine a weight-

less rod in which weights with mass p(x

), i Ú 1, are located at the points x

, i Ú 1

(see Figure 4.4), then the point at which the rod would be in balance is known as the

center of gravity. For those readers acquainted with elementary statics, it is now a

simple matter to show that this point is at E[X].

†

–1 0 2

p(–1) = .10, p(0) = .25, p(2) = .35p(1) = .30,

= center of gravity = .9

FIGURE 4.4

4.4 EXPECTATION OF A FUNCTION OF A RANDOM VARIABLE

Suppose that we are given a discrete random variable along with its probability mass

function and that we want to compute the expected value of some function of X,

say, g(X). How can we accomplish this? One way is as follows: Since g(X) is itself a

discrete random variable, it has a probability mass function, which can be determined

from the probability mass function of X. Once we have determined the probability

mass function of g(X), we can compute E[g(X)] by using the deﬁnition of expected

value.

EXAMPLE 4a

Let X denote a random variable that takes on any of the values −1, 0, and 1 with

respective probabilities

P{X =−1}=.2 P{X = 0}=.5 P{X = 1}=.3

Compute E[X

Solution. Let Y = X

. Then the probability mass function of Y is given by

P{Y = 1}=P{X =−1}+P {X = 1}=.5

P{Y = 0}=P{X = 0}=.5

Hence,

E[X

] = E[Y] = 1(.5) + 0(.5) = .5

†

To prove this, we must show that the sum of the torques tending to turn the point around E[X]

is equal to 0. That is, we must show that 0 =



− E[X])p(x

), which is immediate.

Section 4.4 Expectation of a Function of a Random Variable 129

Note that

.5 = E[X

] Z (E[X])

= .01 .

Although the preceding procedure will always enable us to compute the expected

value of any function of X from a knowledge of the probability mass function of X,

there is another way of thinking about E[g(X)]: Since g(X) will equal g(x) whenever

X is equal to x, it seems reasonable that E[g(X)] should just be a weighted average of

the values g(x),withg(x) being weighted by the probability that X is equal to x. That

is, the following result is quite intuitive:

Proposition 4.1.

If X is a discrete random variable that takes on one of the values x

, i Ú 1, with

respective probabilities p(x

), then, for any real-valued function g,

E[g(X)] =



g(x

)p(x

)

Before proving this proposition, let us check that it is in accord with the results of

Example 4a. Applying it to that example yields

E{X

}=(−1)

(.2) + 0

(.5) + 1

(.3)

= 1(.2 + .3) + 0(.5)

= .5

which is in agreement with the result given in Example 4a.

Proof of Proposition 4.1: The proof of Proposition 4.1 proceeds, as in the preceding

veriﬁcation, by grouping together all the terms in



g(x

)p(x

) having the same value

of g(x

). Speciﬁcally, suppose that y

, j Ú 1, represent the different values of g(x

), i Ú 1.

Then, grouping all the g (x

) having the same value gives



g(x

)p(x

) =



i:g(x

)=y

g(x

)p(x

)



i:g(x

)=y

p(x

)



i:g(x

)=y

p(x

)



P{g(X) = y

}

= E[g(X)]



EXAMPLE 4b

A product that is sold seasonally yields a net proﬁt of b dollars for each unit sold and

a net loss of  dollars for each unit left unsold when the season ends. The number of

units of the product that are ordered at a speciﬁc department store during any season

is a random variable having probability mass function p(i), i Ú 0. If the store must

stock this product in advance, determine the number of units the store should stock

so as to maximize its expected proﬁt.

130 Chapter 4 Random Variables

Solution. Let X denote the number of units ordered. If s units are stocked, then the

proﬁt—call it P(s)—can be expressed as

P(s) = bX − (s − X) if X … s

= sb if X > s

Hence, the expected proﬁt equals

E[P(s)] =



i=0

[bi − (s − i)]p(i) +



i=s+1

sbp(i)

= (b + )



i=0

ip(i) − s



i=0

p(i) + sb

⎡

⎣

1 −



i=0

p(i)

⎤

⎦

= (b + )



i=0

ip(i) − (b + )s



i=0

p(i) + sb

= sb + (b + )



i=0

(i − s)p(i)

To determine the optimum value of s, let us investigate what happens to the proﬁt

when we increase s by 1 unit. By substitution, we see that the expected proﬁt in this

case is given by

E[P(s + 1)] = b(s + 1) + (b + )

s+1



i=0

(i − s − 1)p(i)

= b(s + 1) + (b + )



i=0

(i − s − 1)p(i)

Therefore,

E[P(s + 1)] − E[P(s)] = b − (b + )



i=0

p(i)

Thus, stocking s + 1 units will be better than stocking s units whenever



i=0

p(i)<

b + 

(4.1)

Because the left-hand side of Equation (4.1) is increasing in s while the right-hand

side is constant, the inequality will be satisﬁed for all values of s … s

∗

, where s

∗

is the

largest value of s satisfying Equation (4.1). Since

E[P(0)] < ··· < E[P(s

∗

)] < E[P(s

∗

+ 1)] > E[P(s

∗

+ 2)] > ···

it follows that stocking s

∗

+ 1 items will lead to a maximum expected proﬁt. .

EXAMPLE 4c Utility

Suppose that you must choose one of two possible actions, each of which can result

in any of n consequences, denoted as C

, ..., C

. Suppose that if the ﬁrst action is

Section 4.4 Expectation of a Function of a Random Variable 131

chosen, then consequence C

will result with probability p

, i = 1, ..., n, whereas

if the second action is chosen, then consequence C

will result with probability q

i = 1, ..., n, where



i=1



i=1

= 1. The following approach can be used to deter-

mine which action to choose: Start by assigning numerical values to the different

consequences in the following manner: First, identify the least and the most desir-

able consequences—call them c and C, respectively; give consequence c the value 0

and give C the value 1. Now consider any of the other n − 2 consequences, say, C

.To

value this consequence, imagine that you are given the choice between either receiv-

ing C

or taking part in a random experiment that either earns you consequence C

with probability u or consequence c with probability 1 − u. Clearly, your choice will

depend on the value of u. On the one hand, if u = 1, then the experiment is certain

to result in consequence C, and since C is the most desirable consequence, you will

prefer participating in the experiment to receiving C

. On the other hand, if u = 0,

then the experiment will result in the least desirable consequence—namely, c—so in

this case you will prefer the consequence C

to participating in the experiment. Now,

as u decreases from 1 to 0, it seems reasonable that your choice will at some point

switch from participating in the experiment to the certain return of C

, and at that

critical switch point you will be indifferent between the two alternatives. Take that

indifference probability u as the value of the consequence C

. In other words, the

value of C

is that probability u such that you are indifferent between either receiv-

ing the consequence C

or taking part in an experiment that returns consequence C

with probability u or consequence c with probability 1 − u. We call this indifference

probability the utility of the consequence C

, and we designate it as u(C

To determine which action is superior, we need to evaluate each one. Consider the

ﬁrst action, which results in consequence C

with probability p

, i = 1, ..., n. We can

think of the result of this action as being determined by a two-stage experiment. In the

ﬁrst stage, one of the values 1, ..., n is chosen according to the probabilities p

, ..., p

;

if value i is chosen, you receive consequence C

. However, since C

is equivalent to

obtaining consequence C with probability u(C

) or consequence c with probability

1 − u(C

), it follows that the result of the two-stage experiment is equivalent to an

experiment in which either consequence C or consequence c is obtained, with C being

obtained with probability



i=1

u(C

)

Similarly, the result of choosing the second action is equivalent to taking part in an

experiment in which either consequence C or consequence c is obtained, with C being

obtained with probability



i=1

u(C

)

Since C is preferable to c, it follows that the ﬁrst action is preferable to the second

action if



i=1

u(C



i=1

u(C

)

In other words, the worth of an action can be measured by the expected value of the

utility of its consequence, and the action with the largest expected utility is the most

preferable. .

132 Chapter 4 Random Variables

A simple logical consequence of Proposition 4.1 is Corollary 4.1.

Corollary 4.1. If a and b are constants, then

E[aX + b] = aE[X] + b

Proof.

E[aX + b] =



x:p(x)>0

(ax + b)p(x)

= a



x:p(x)>0

xp(x) + b



x:p(x)>0

p(x)

= aE[X] + b

The expected value of a random variable X, E[X], is also referred to as the mean

or the ﬁrst moment of X. The quantity E[X

], n Ú 1, is called the nth moment of X.

By Proposition 4.1, we note that

E[X

] =



x:p(x)>0

p(x)

4.5 VARIANCE

Given a random variable X along with its distribution function F, it would be

extremely useful if we were able to summarize the essential properties of F by cer-

tain suitably deﬁned measures. One such measure would be E[X], the expected value

of X. However, although E[X] yields the weighted average of the possible values of

X, it does not tell us anything about the variation, or spread, of these values. For

instance, although random variables W, Y,andZ having probability mass functions

determined by

W = 0 with probability 1

Y =

⎧

⎨

⎩

−1 with probability

+1 with probability

Z =

⎧

⎨

⎩

−100 with probability

+100 with probability

all have the same expectation—namely, 0—there is a much greater spread in the pos-

sible values of Y than in those of W (which is a constant) and in the possible values

of Z than in those of Y.

Because we expect X to take on values around its mean E[X], it would appear that

a reasonable way of measuring the possible variation of X would be to look at how

far apart X would be from its mean, on the average. One possible way to measure this

variation would be to consider the quantity E[|X − μ|], where μ = E[X]. However,

it turns out to be mathematically inconvenient to deal with this quantity, so a more

tractable quantity is usually considered—namely, the expectation of the square of the

difference between X and its mean. We thus have the following deﬁnition.

Section 4.5 Variance 133

Deﬁnition

If X is a random variable with mean μ, then the variance of X, denoted by Var(X),

is deﬁned by

Var(X) = E[(X − μ)

]

An alternative formula for Var(X) is derived as follows:

Var(X) = E[(X − μ)

]



(x − μ)

p(x)



− 2μx + μ

)p(x)



p(x) − 2μ



xp(x) + μ



p(x)

= E[X

] − 2μ

+ μ

= E[X

] − μ

That is,

Var(X) = E[X

] − (E[X])

In words, the variance of X is equal to the expected value of X

minus the square

of its expected value. In practice, this formula frequently offers the easiest way to

compute Var(X).

EXAMPLE 5a

Calculate Var(X) if X represents the outcome when a fair die is rolled.

Solution. It was shown in Example 3a that E[X] =

. Also,

E[X

] = 1





+ 2





+ 3





+ 4





+ 5





+ 6









(91)

Hence,

Var(X) =

−





A useful identity is that, for any constants a and b,

Var(aX + b) = a

Var(X)

134 Chapter 4 Random Variables

To prove this equality, let μ = E[X] and note from Corollary 4.1 that E [aX + b] =

aμ + b. Therefore,

Var(aX + b) = E[(aX + b − aμ − b)

]

= E[a

(X − μ)

]

= a

E[(X − μ)

]

= a

Var(X)

Remarks. (a) Analogous to the means being the center of gravity of a distribu-

tion of mass, the variance represents, in the terminology of mechanics, the moment

of inertia.

(b) The square root of the Var(X) is called the standard deviation of X, and we

denote it by SD(X). That is,

SD(X) =

Var(X)

Discrete random variables are often classiﬁed according to their probability mass

functions. In the next few sections, we consider some of the more common types.

4.6 THE BERNOULLI AND BINOMIAL RANDOM VARIABLES

Suppose that a trial, or an experiment, whose outcome can be classiﬁed as either a

success or a failure is performed. If we let X = 1 when the outcome is a success and

X = 0 when it is a failure, then the probability mass function of X is given by

p(0) = P{X = 0}=1 − p

p(1) = P{X = 1}=p

(6.1)

where p,0 … p … 1, is the probability that the trial is a success.

A random variable X is said to be a Bernoulli random variable (after the Swiss

mathematician James Bernoulli) if its probability mass function is given by Equa-

tions (6.1) for some p ∈ (0, 1).

Suppose now that n independent trials, each of which results in a success with prob-

ability p and in a failure with probability 1 − p, are to be performed. If X represents

the number of successes that occur in the n trials, then X is said to be a binomial

random variable with parameters (n, p). Thus, a Bernoulli random variable is just a

binomial random variable with parameters (1, p).

The probability mass function of a binomial random variable having parameters

(n, p) is given by

i) =





(1 − p)

n−i

i = 0, 1, ..., n (6.2)

The validity of Equation (6.2) may be veriﬁed by ﬁrst noting that the probability of

any particular sequence of n outcomes containing i successes and n − i failures is,

by the assumed independence of trials, p

(1 − p)

n−i

. Equation (6.2) then follows,

since there are





different sequences of the n outcomes leading to i successes and

n − i failures. This perhaps can most easily be seen by noting that there are





different choices of the i trials that result in successes. For instance, if n = 4, i = 2,

then there are





= 6 ways in which the four trials can result in two successes,

Section 4.6 The Bernoulli and Binomial Random Variables 135

namely, any of the outcomes (s, s, f , f ), (s, f , s, f ), (s, f , f , s), (f , s, s, f ), (f , s, f , s), and

(f , f , s, s), where the outcome (s, s, f , f ) means, for instance, that the ﬁrst two trials

are successes and the last two failures. Since each of these outcomes has probability

(1 − p)

of occurring, the desired probability of two successes in the four trials





(1 − p)

Note that, by the binomial theorem, the probabilities sum to 1; that is,



i=0

p(i) =



i=0





(1 − p)

n−i

= [p + (1 − p)]

= 1

EXAMPLE 6a

Five fair coins are ﬂipped. If the outcomes are assumed independent, ﬁnd the proba-

bility mass function of the number of heads obtained.

Solution. If we let X equal the number of heads (successes) that appear, then X

is a binomial random variable with parameters



n = 5, p =



. Hence, by Equa-

tion (6.2),

P{X = 0}=











P{X = 1}=











P{X = 2}=











P{X = 3}=











P{X = 4}=











P{X = 5}=











EXAMPLE 6b

It is known that screws produced by a certain company will be defective with proba-

bility .01, independently of each other. The company sells the screws in packages of

10 and offers a money-back guarantee that at most 1 of the 10 screws is defective.

What proportion of packages sold must the company replace?

Solution. If X is the number of defective screws in a package, then X is a binomial

random variable with parameters (10, .01). Hence, the probability that a package will

have to be replaced is

1 − P {X = 0}−P{X = 1}=1 −





(.01)

(.99)

−





(.01)

(.99)

L .004

Thus, only .4 percent of the packages will have to be replaced. .

136 Chapter 4 Random Variables

EXAMPLE 6c

The following gambling game, known as the wheel of fortune (or chuck-a-luck), is

quite popular at many carnivals and gambling casinos: A player bets on one of the

numbers 1 through 6. Three dice are then rolled, and if the number bet by the player

appears i times, i = 1, 2, 3, then the player wins i units; if the number bet by the player

does not appear on any of the dice, then the player loses 1 unit. Is this game fair to

the player? (Actually, the game is played by spinning a wheel that comes to rest on

a slot labeled by three of the numbers 1 through 6, but this variant is mathematically

equivalent to the dice version.)

Solution. If we assume that the dice are fair and act independently of each other,

then the number of times that the number bet appears is a binomial random variable

with parameters





. Hence, letting X denote the player’s winnings in the game,

we have

P{X =−1}=











125

216

P{X = 1}=











216

P{X = 2}=











216

P{X = 3}=











216

In order to determine whether or not this is a fair game for the player, let us calcu-

late E[X]. From the preceding probabilities, we obtain

E[X] =

−125 + 75 + 30 + 3

216

−17

216

Hence, in the long run, the player will lose 17 units per every 216 games he plays. .

In the next example, we consider the simplest form of the theory of inheritance as

developed by Gregor Mendel (1822–1884).

EXAMPLE 6d

Suppose that a particular trait (such as eye color or left-handedness) of a person is

classiﬁed on the basis of one pair of genes, and suppose also that d represents a dom-

inant gene and r a recessive gene. Thus, a person with dd genes is purely dominant,

one with rr is purely recessive, and one with rd is hybrid. The purely dominant and the

hybrid individuals are alike in appearance. Children receive 1 gene from each parent.

If, with respect to a particular trait, 2 hybrid parents have a total of 4 children, what is

the probability that 3 of the 4 children have the outward appearance of the dominant

gene?

Solution. If we assume that each child is equally likely to inherit either of 2 genes

from each parent, the probabilities that the child of 2 hybrid parents will have dd,

rr,andrd pairs of genes are, respectively,

,and

. Hence, since an offspring will

Section 4.6 The Bernoulli and Binomial Random Variables 137

have the outward appearance of the dominant gene if its gene pair is either dd or rd,

it follows that the number of such children is binomially distributed with parameters





. Thus, the desired probability is











EXAMPLE 6e

Consider a jury trial in which it takes 8 of the 12 jurors to convict the defendant; that

is, in order for the defendant to be convicted, at least 8 of the jurors must vote him

guilty. If we assume that jurors act independently and that, whether or not the defen-

dant is guilty, each makes the right decision with probability θ, what is the probability

that the jury renders a correct decision?

Solution. The problem, as stated, is incapable of solution, for there is not yet enough

information. For instance, if the defendant is innocent, the probability of the jury’s

rendering a correct decision is



i=5





(1 − θ)

12−i

whereas, if he is guilty, the probability of a correct decision is



i=8





(1 − θ)

12−i

Therefore, if α represents the probability that the defendant is guilty, then, by condi-

tioning on whether or not he is guilty, we obtain the probability that the jury renders

a correct decision:



i=8





(1 − θ)

12−i

+ (1 − α)



i=5





(1 − θ)

12−i

EXAMPLE 6f

A communication system consists of n components, each of which will, indepen-

dently, function with probability p. The total system will be able to operate effectively

if at least one-half of its components function.

(a) For what values of p is a 5-component system more likely to operate effectively

than a 3-component system?

(b) In general, when is a (2k + 1)-component system better than a (2k − 1)-

component system?

Solution. (a) Because the number of functioning components is a binomial random

variable with parameters (n, p), it follows that the probability that a 5-component

system will be effective is





(1 − p)





(1 − p) + p

138 Chapter 4 Random Variables

whereas the corresponding probability for a 3-component system is





(1 − p) + p

Hence, the 5-component system is better if

10p

(1 − p)

+ 5p

(1 − p) + p

> 3p

(1 − p) + p

which reduces to

3(p − 1)

(2p − 1)>0

p >

(b) In general, a system with 2k + 1 components will be better than one with

2k − 1 components if (and only if) p >

. To prove this, consider a system of 2k + 1

components and let X denote the number of the ﬁrst 2k − 1 that function. Then

2k+1

(effective) = P{X Ú k + 1}+P{X = k}(1 − (1 − p)

)

+ P{X = k − 1}p

which follows because the (2k + 1)-component system will be effective if either

(i) X Ú k + 1;

(ii) X = k and at least one of the remaining 2 components function; or

(iii) X = k − 1 and both of the next 2 components function.

Since

2k−1

(effective) = P{X Ú k}

= P{X = k}+P{X Ú k + 1}

we obtain

2k+1

(effective) − P

2k−1

(effective)

= P{X = k − 1}p

− (1 − p)

P{X = k}



2k − 1

k − 1



k−1

(1 − p)

− (1 − p)



2k − 1



(1 − p)

k−1



2k − 1



(1 − p)

[ p − (1 − p)] since



2k − 1

k − 1





2k − 1



> 0 3 p >

Section 4.6 The Bernoulli and Binomial Random Variables 139

4.6.1 Properties of Binomial Random Variables

We will now examine the properties of a binomial random variable with parameters

n and p. To begin, let us compute its expected value and variance. Now,

E[X

] =



i=0





(1 − p)

n−i



i=1





(1 − p)

n−i

Using the identity





= n



n − 1

i − 1



gives

E[X

] = np



i=1

k−1



n − 1

i − 1



i−1

(1 − p)

n−i

= np

n−1



j=0

(j + 1)

k−1



n − 1



(1 − p)

n−1−j

by letting

j = i − 1

= npE[(Y + 1)

k−1

]

where Y is a binomial random variable with parameters n − 1, p. Setting k = 1inthe

preceding equation yields

E[X] = np

That is, the expected number of successes that occur in n independent trials when

each is a success with probability p is equal to np. Setting k = 2 in the preceding equa-

tion, and using the preceding formula for the expected value of a binomial random

variable yields

E[X

] = npE[Y + 1]

= np[(n − 1)p + 1]

Since E[X] = np, we obtain

Var(X) = E[X

] − (E[X])

= np[(n − 1)p + 1] − (np)

= np(1 − p)

Summing up, we have shown the following:

If X is a binomial random variable with parameters n and p,then

E[X] = np

Var(X) = np(1 − p)

140 Chapter 4 Random Variables

The following proposition details how the binomial probability mass function ﬁrst

increases and then decreases.

Proposition 6.1. If X is a binomial random variable with parameters (n, p), where

0 < p < 1, then as k goes from 0 to n, P{X = k} ﬁrst increases monotonically and

then decreases monotonically, reaching its largest value when k is the largest integer

less than or equal to (n + 1)p.

Proof. We prove the proposition by considering P{X = k}/P{X = k − 1} and

determining for what values of k it is greater or less than 1. Now,

P{X = k}

P{X = k − 1}

(n − k)!k!

(1 − p)

n−k

(n − k + 1)!(k − 1)!

k−1

(1 − p)

n−k+1

(n − k + 1)p

k(1 − p)

Hence, P{X = k} Ú P{X = k − 1} if and only if

(n − k + 1)p Ú k(1 − p)

or, equivalently, if and only if

k … (n + 1)p

and the proposition is proved. .

As an illustration of Proposition 6.1 consider Figure 4.5, the graph of the probabil-

ity mass function of a binomial random variable with parameters (10,

EXAMPLE 6g

In a U.S. presidential election, the candidate who gains the maximum number of

votes in a state is awarded the total number of electoral college votes allocated to

1024 ⫻ p(k)

1234 567 8 9 10

120

210

252

FIGURE 4.5 Graph of p(k) =









Section 4.6 The Bernoulli and Binomial Random Variables 141

that state. The number of electoral college votes of a given state is roughly propor-

tional to the population of that state—that is, a state with population n has roughly

nc electoral votes. (Actually, it is closer to nc + 2, as a state is given an electoral

vote for each member it has in the House of Representatives, with the number of

such representatives being roughly proportional to the population of the state, and

one electoral college vote for each of its two senators.) Let us determine the average

power of a citizen in a state of size n in a close presidential election, where, by average

power in a close election, we mean that a voter in a state of size n = 2k + 1 will be

decisive if the other n − 1 voters split their votes evenly between the two candidates.

(We are assuming here that n is odd, but the case where n is even is quite similar.)

Because the election is close, we shall suppose that each of the other n − 1 = 2k

voters acts independently and is equally likely to vote for either candidate. Hence,

the probability that a voter in a state of size n = 2k + 1 will make a difference to

the outcome is the same as the probability that 2k tosses of a fair coin land heads and

tails an equal number of times. That is,

P{voter in state of size 2k + 1 makes a difference}











(2k)!

k!k!2

To approximate the preceding equality, we make use of Stirling’s approximation,

which says that, for k large,

k! ∼ k

k+1/2

−k

√

2π

where we say that a

∼ b

when the ratio a

approaches 1 as k approaches q.

Hence, it follows that

P{voter in state of size 2k + 1 makes a difference}

∼

(2k)

2k+1/2

−2k

√

2π

2k+1

−2k

(2π)2

√

kπ

Because such a voter (if he or she makes a difference) will affect nc electoral votes,

the expected number of electoral votes a voter in a state of size n will affect—or the

voter’s average power—is given by

average power = ncP{makes a difference}

∼

nπ/2

= c

2n/π

Thus, the average power of a voter in a state of size n is proportional to the square

root of n, showing that, in presidential elections, voters in large states have more

power than do those in smaller states. .

142 Chapter 4 Random Variables

4.6.2 Computing the Binomial Distribution Function

Suppose that X is binomial with parameters (n, p). The key to computing its distribu-

tion function

P{X … i}=



k=0





(1 − p)

n−k

i = 0, 1, ..., n

is to utilize the following relationship between P{X = k + 1} and P{X = k}, which

was established in the proof of Proposition 6.1:

P{X = k + 1}=

1 − p

n − k

k + 1

P{X = k} (6.3)

EXAMPLE 6h

Let X be a binomial random variable with parameters n = 6, p = .4. Then, starting

with P{X = 0}=(.6)

and recursively employing Equation (6.3), we obtain

P{X = 0}=(.6)

L .0467

P{X = 1}=

P{X = 0} L .1866

P{X = 2}=

P{X = 1} L .3110

P{X = 3}=

P{X = 2} L .2765

P{X = 4}=

P{X = 3} L .1382

P{X = 5}=

P{X = 4} L .0369

P{X = 6}=

P{X = 5} L .0041

A computer program that utilizes the recursion (6.3) to compute the binomial

distribution function is easily written. To compute P{X … i}, the program should

ﬁrst compute P{X = i} and then use the recursion to successively compute P{X =

i − 1}, P{X = i − 2}, and so on.

Historical Note

Independent trials having a common probability of success p were ﬁrst studied

by the Swiss mathematician Jacques Bernoulli (1654–1705). In his book Ars Con-

jectandi (The Art of Conjecturing), published by his nephew Nicholas eight years

after his death in 1713, Bernoulli showed that if the number of such trials were

large, then the proportion of them that were successes would be close to p with a

probability near 1.

Jacques Bernoulli was from the ﬁrst generation of the most famous mathemat-

ical family of all time. Altogether, there were between 8 and 12 Bernoullis, spread

over three generations, who made fundamental contributions to probability, statis-

tics, and mathematics. One difﬁculty in knowing their exact number is the fact that

several had the same name. (For example, two of the sons of Jacques’s brother Jean

Section 4.7 The Poisson Random Variable 143

were named Jacques and Jean.) Another difﬁculty is that several of the Bernoullis

were known by different names in different places. Our Jacques (sometimes writ-

ten Jaques) was, for instance, also known as Jakob (sometimes written Jacob) and

as James Bernoulli. But whatever their number, their inﬂuence and output were

prodigious. Like the Bachs of music, the Bernoullis of mathematics were a family

for the ages!

EXAMPLE 6i

If X is a binomial random variable with parameters n = 100 and p = .75, ﬁnd P{X =

70} and P{X … 70}.

Solution. The answer is shown here in Figure 4.6.

Enter Value For p:.75

Enter Value For n:

100

Enter Value For i:

Binomial Distribution

Start

Quit

Probability (Number of Successes = i) = .04575381

Probability (Number of Successes < = i) = .14954105

FIGURE 4.6

4.7 THE POISSON RANDOM VARIABLE

A random variable X that takes on one of the values 0, 1, 2, ... is said to be a Poisson

random variable with parameter λ if, for some λ>0,

p(i) = P{X = i}=e

−λ

i = 0, 1, 2, ... (7.1)

Equation (7.1) deﬁnes a probability mass function, since



i=0

p(i) = e

−λ



i=0

= e

−λ

= 1

The Poisson probability distribution was introduced by Sim

eon Denis Poisson in a

book he wrote regarding the application of probability theory to lawsuits, criminal

trials, and the like. This book, published in 1837, was entitled Recherches sur la prob-

abilit

e des jugements en mati

ere criminelle et en mati

ere civile (Investigations into the

Probability of Verdicts in Criminal and Civil Matters).

The Poisson random variable has a tremendous range of applications in diverse

areas because it may be used as an approximation for a binomial random variable

with parameters (n, p) when n is large and p is small enough so that np is of moderate

144 Chapter 4 Random Variables

size. To see this, suppose that X is a binomial random variable with parameters (n, p),

and let λ = np. Then

P{X = i}=

(n − i)!i!

(1 − p)

n−i

(n − i)!i!







1 −



n−i

n(n − 1) ···(n − i + 1)

(1 − λ/n)

Now, for n large and λ moderate,



1 −



L e

−λ

n(n − 1) ···(n − i + 1)

L 1



1 −



L 1

Hence, for n large and λ moderate,

P{X = i} L e

−λ

In other words, if n independent trials, each of which results in a success with

probability p, are performed, then, when n is large and p is small enough to make

np moderate, the number of successes occurring is approximately a Poisson random

variable with parameter λ = np. This value λ (which will later be shown to equal the

expected number of successes) will usually be determined empirically.

Some examples of random variables that generally obey the Poisson probability

law [that is, they obey Equation (7.1)] are as follows:

1. The number of misprints on a page (or a group of pages) of a book

2. The number of people in a community who survive to age 100

3. The number of wrong telephone numbers that are dialed in a day

4. The number of packages of dog biscuits sold in a particular store each day

5. The number of customers entering a post ofﬁce on a given day

6. The number of vacancies occurring during a year in the federal judicial system

7. The number of α-particles discharged in a ﬁxed period of time from some radioac-

tive material

Each of the preceding, and numerous other random variables, are approximately

Poisson for the same reason—namely, because of the Poisson approximation to the

binomial. For instance, we can suppose that there is a small probability p that each

letter typed on a page will be misprinted. Hence, the number of misprints on a page

will be approximately Poisson with λ = np, where n is the number of letters on a

page. Similarly, we can suppose that each person in a community has some small

probability of reaching age 100. Also, each person entering a store may be thought of

as having some small probability of buying a package of dog biscuits, and so on.

EXAMPLE 7a

Suppose that the number of typographical errors on a single page of this book has a

Poisson distribution with parameter λ =

. Calculate the probability that there is at

least one error on this page.

Section 4.7 The Poisson Random Variable 145

Solution. Letting X denote the number of errors on this page, we have

P{X Ú 1}=1 − P{X = 0}=1 − e

−1/2

L .393

EXAMPLE 7b

Suppose that the probability that an item produced by a certain machine will be

defective is .1. Find the probability that a sample of 10 items will contain at most

1 defective item.

Solution. The desired probability is





(.1)

(.9)





(.1)

(.9)

= .7361,

whereas the Poisson approximation yields the value e

−1

+ e

−1

L .7358. .

EXAMPLE 7c

Consider an experiment that consists of counting the number of α particles given

off in a 1-second interval by 1 gram of radioactive material. If we know from past

experience that, on the average, 3.2 such α particles are given off, what is a good

approximation to the probability that no more than 2 α particles will appear?

Solution. If we think of the gram of radioactive material as consisting of a large num-

ber n of atoms, each of which has probability of 3.2/n of disintegrating and sending off

an α particle during the second considered, then we see that, to a very close approx-

imation, the number of α particles given off will be a Poisson random variable with

parameter λ = 3.2. Hence, the desired probability is

P{X … 2}=e

−3.2

+ 3.2e

−3.2

(3.2)

−3.2

L .3799

Before computing the expected value and variance of the Poisson random variable

with parameter λ, recall that this random variable approximates a binomial random

variable with parameters n and p when n is large, p is small, and λ = np. Since such

a binomial random variable has expected value np = λ and variance np(1 − p) =

λ(1 − p) L λ (since p is small), it would seem that both the expected value and the

variance of a Poisson random variable would equal its parameter λ. We now verify

this result:

E[X] =



i=0

−λ

= λ



i=1

−λ

i−1

(i − 1)!

= λe

−λ



j=0

by letting

j = i − 1

= λ since



j=0

= e

146 Chapter 4 Random Variables

Thus, the expected value of a Poisson random variable X is indeed equal to its

parameter λ. To determine its variance, we ﬁrst compute E[X

E[X

] =



i=0

−λ

= λ



i=1

−λ

i−1

(i − 1)!

= λ



j=0

(j + 1)e

−λ

by letting

j = i − 1

= λ

⎡

⎢

⎣



j=0

−λ



j=0

−λ

⎤

⎥

⎦

= λ(λ + 1)

where the ﬁnal equality follows because the ﬁrst sum is the expected value of a Pois-

son random variable with parameter λ and the second is the sum of the probabilities

of this random variable. Therefore, since we have shown that E[X] = λ, we obtain

Var(X) = E[X

] − (E[X])

= λ

Hence, the expected value and variance of a Poisson random variable are both

equal to its parameter λ.

We have shown that the Poisson distribution with parameter np is a very good

approximation to the distribution of the number of successes in n independent trials

when each trial has probability p of being a success, provided that n is large and p

small. In fact, it remains a good approximation even when the trials are not inde-

pendent, provided that their dependence is weak. For instance, recall the matching

problem (Example 5m of Chapter 2) in which n men randomly select hats from a set

consisting of one hat from each person. From the point of view of the number of men

who select their own hat, we may regard the random selection as the result of n tri-

als where we say that trial i is a success if person i selects his own hat, i = 1, ..., n.

Deﬁning the events E

, i = 1, ..., n,by

={trial i is a success}

it is easy to see that

P{E

and P{E

n − 1

, j Z i

Thus, we see that, although the events E

, i = 1, ..., n are not independent, their

dependence, for large n, appears to be weak. Because of this it seems reasonable to

expect that the number of successes will approximately have a Poisson distribution

with parameter n * 1/n = 1, and indeed this is veriﬁed in Example 5m of Chapter 2.

For a second illustration of the strength of the Poisson approximation when the

trials are weakly dependent, let us consider again the birthday problem presented in

Example 5i of Chapter 2. In this example, we suppose that each of n people is equally

likely to have any of the 365 days of the year as his or her birthday, and the problem

Section 4.7 The Poisson Random Variable 147

is to determine the probability that a set of n independent people all have different

birthdays. A combinatorial argument was used to determine this probability, which

was shown to be less than

when n = 23.

We can approximate the preceding probability by using the Poisson approximation

as follows: Imagine that we have a trial for each of the





pairs of individuals i and

j, i Z j, and say that trial i, j is a success if persons i and j have the same birthday. If

we let E

denote the event that trial i, j is a success, then, whereas the





events

,1 … i < j … n, are not independent (see Theoretical Exercise 21), their depen-

dence appears to be rather weak. (Indeed, these events are even pairwise indepen-

dent, in that any 2 of the events E

and E

are independent—again, see Theoretical

Exercise 21). Since P(E

) = 1/365, it is reasonable to suppose that the number of

successes should approximately have a Poisson distribution with mean







365 =

n(n − 1)/730. Therefore,

P{no 2 people have the same birthday}=P{0 successes}

L exp

−n(n − 1)

730

To determine the smallest integer n for which this probability is less than

, note that

exp

−n(n − 1)

730

…

is equivalent to

exp

n(n − 1)

730

Ú 2

Taking logarithms of both sides, we obtain

n(n − 1) Ú 730 log 2

L 505.997

which yields the solution n = 23, in agreement with the result of Example 5i of

Chapter 2.

Suppose now that we wanted the probability that, among the n people, no 3 of

them have their birthday on the same day. Whereas this now becomes a difﬁcult com-

binatorial problem, it is a simple matter to obtain a good approximation. To begin,

imagine that we have a trial for each of the





triplets i, j, k, where 1 … i < j < k … n,

and call the i, j, k trial a success if persons i, j,andk all have their birthday on the same

day. As before, we can then conclude that the number of successes is approximately

a Poisson random variable with parameter





P{i, j, k have the same birthday}=





365



n(n − 1)(n − 2)

6 * (365)

148 Chapter 4 Random Variables

Hence,

P{no 3 have the same birthday} L exp

−n(n − 1)(n − 2)

799350

This probability will be less than

when n is such that

n(n − 1)(n − 2) Ú 799350 log 2 L 554067.1

which is equivalent to n Ú 84. Thus, the approximate probability that at least 3 people

in a group of size 84 or larger will have the same birthday exceeds

For the number of events to occur to approximately have a Poisson distribution, it

is not essential that all the events have the same probability of occurrence, but only

that all of these probabilities be small. The following is referred to as the Poisson

paradigm.

Poisson Paradigm. Consider n events, with p

equal to the probability that event i

occurs, i = 1, ..., n. If all the p

are “small” and the trials are either independent or at

most “weakly dependent,” then the number of these events that occur approximately

has a Poisson distribution with mean



i=1

Our next example not only makes use of the Poisson paradigm, but also illustrates

a variety of the techniques we have studied so far.

EXAMPLE 7d Length of the longest run

A coin is ﬂipped n times. Assuming that the ﬂips are independent, with each one

coming up heads with probability p, what is the probability that there is a string of k

consecutive heads?

Solution. We will ﬁrst use the Poisson paradigm to approximate this probability.

Now, if, for i = 1, ..., n − k + 1, we let H

denote the event that ﬂips i, i + 1, ..., i +

k − 1 all land on heads, then the desired probability is that at least one of the events

occur. Because H

is the event that, starting with ﬂip i, the next k ﬂips all land on

heads, it follows that P(H

) = p

. Thus, when p

is small, we might think that the num-

ber of the H

that occur should have an approximate Poisson distribution. However,

such is not the case, because, although the events all have small probabilities, some of

their dependencies are too great for the Poisson distribution to be a good approxima-

tion. For instance, because the conditional probability that ﬂips 2, ..., k + 1 are all

heads given that ﬂips 1, ..., k are all heads is equal to the probability that ﬂip k + 1

is a head, it follows that

P(H

) = p

which is far greater than the unconditional probability of H

The trick that enables us to use a Poisson approximation is to note that there

will be a string of k consecutive heads either if there is such a string that is imme-

diately followed by a tail or if the ﬁnal k ﬂips all land on heads. Consequently, for

i = 1, ..., n − k, let E

be the event that ﬂips i, ..., i + k − 1 are all heads and ﬂip

i + k is a tail; also, let E

n−k+1

be the event that ﬂips n − k + 1, ..., n are all heads.

Note that

P(E

) = p

(1 − p), i … n − k

P(E

n−k+1

) = p

Section 4.7 The Poisson Random Variable 149

Thus, when p

is small, each of the events E

has a small probability of occurring.

Moreover, for i Z j, if the events E

and E

refer to nonoverlapping sequences of ﬂips,

then P(E

) = P(E

); if they refer to overlapping sequences, then P(E

) = 0.

Hence, in both cases, the conditional probabilities are close to the unconditional ones,

indicating that N, the number of the events E

that occur, should have an approximate

Poisson distribution with mean

E[N] =

n−k+1



i=1

P(E

) = (n − k)p

(1 − p) + p

Because there will not be a run of k heads if (and only if) N = 0, thus the preceding

gives

P(no head strings of length k) = P(N = 0) L exp{−(n − k)p

(1 − p) − p

}

If we let L

denote the largest number of consecutive heads in the n ﬂips, then,

because L

will be less than k if (and only if) there are no head strings of length

k, the preceding equation can be written as

P{L

< k} L exp{−(n − k)p

(1 − p) − p

}

Now, let us suppose that the coin being ﬂipped is fair; that is, suppose that p = 1/2.

Then the preceding gives

P{L

< k} L exp

−

n − k + 2

k+1

L exp

−

k+1

where the ﬁnal approximation supposes that e

k−2

k+1

L 1 (that is, that

k−2

k+1

L 0). Let

j = log

n, and assume that j is an integer. For k = j + i,

k+1

i+1

Consequently,

P{L

< j + i} L exp{−(1/2)

i+1

}

which implies that

P{L

= j + i}=P{L

< j + i + 1}−P{L

< j + i}

L exp{−(1/2)

i+2

}−exp{−(1/2)

i+1

}

For instance,

P{L

< j − 3} L e

−4

L .0183

P{L

= j − 3} L e

−2

− e

−4

L .1170

P{L

= j − 2} L e

−1

− e

−2

L .2325

P{L

= j − 1} L e

−1/2

− e

−1

L .2387

P{L

= j} L e

−1/4

− e

−1/2

L .1723

P{L

= j + 1} L e

−1/8

− e

−1/4

L .1037

P{L

= j + 2} L e

−1/16

− e

−1/8

L .0569

P{L

= j + 3} L e

−1/32

− e

−1/16

L .0298

P{L

Ú j + 4} L 1 − e

−1/32

L .0308

150 Chapter 4 Random Variables

Thus, we observe the rather interesting fact that no matter how large n is, the length

of the longest run of heads in a sequence of n ﬂips of a fair coin will be within 2 of

log

(n) − 1 with a probability approximately equal to .86.

We now derive an exact expression for the probability that there is a string of

k consecutive heads when a coin that lands on heads with probability p is ﬂipped

n times. With the events E

, i = 1, ..., n − k + 1, as deﬁned earlier, and with L

denoting, as before, the length of the longest run of heads,

P(L

Ú k) = P(there is a string of k consecutive heads) = P(∪

n−k+1

i=1

)

The inclusion–exclusion identity for the probability of a union can be written as

P(∪

n−k+1

i=1

) =

n−k+1



r=1

(−1)

r+1



<···<i

P(E

···E

)

Let S

denote the set of ﬂip numbers to which the event E

refers. (So, for instance,

={1, ..., k + 1}.) Now, consider one of the r-way intersection probabilities that

does not include the event E

n−k+1

. That is, consider P(E

···E

) where i

< ··· <

< n − k + 1. On the one hand, if there is any overlap in the sets S

, ..., S

then this

probability is 0. On the other hand, if there is no overlap, then the events E

, ..., E

are independent. Therefore,

P(E

···E

) =

0, if there is any overlap in S

, ..., S

(1 − p)

, if there is no overlap

We must now determine the number of different choices of i

< ···< i

< n − k + 1

for which there is no overlap in the sets S

, ..., S

. To do so, note ﬁrst that each of

the S

, j = 1, ..., r, refer to k + 1 ﬂips, so, without any overlap, they together refer

to r(k + 1) ﬂips. Now consider any permutation of r identical letters a (one for each

of the sets S

, ..., S

r−1

) and of n − r(k + 1) identical letters b (one for each of the

trials that are not part of any of S

, ..., S

r−1

, S

n−k+1

). Interpret the number of b’s

before the ﬁrst a as the number of ﬂips before S

, the number of b’s between the ﬁrst

and second a as the number of ﬂips between S

and S

, and so on, with the number

of b’s after the ﬁnal a representing the number of ﬂips after S

. Because there are



n−rk



permutations of r letters a and of n − r(k + 1) letters b, with every such

permutation corresponding (in a one-to-one fashion) to a different nonoverlapping

choice, it follows that



<···<i

<n−k+1

P(E

···E

) =



n − rk



(1 − p)

We must now consider r-way intersection probabilities of the form

P(E

···E

r−1

n−k+1

where i

< ··· < i

r−1

< n − k + 1. Now, this probability will equal 0 if there is any

overlap in S

, ..., S

r−1

, S

n−k

; if there is no overlap, then the events of the intersection

will be independent, so

P(E

···E

r−1

n−k+1

) = [p

(1 − p)]

r−1

= p

(1 − p)

r−1

By a similar argument as before, the number of nonoverlapping sets S

, ..., S

r−1

, S

n−k

will equal the number of permutations of r − 1 letters a (one for each of the sets

Section 4.7 The Poisson Random Variable 151

, ..., S

r−1

) and of n − (r − 1)(k + 1) − k = n − rk − (r − 1) letters b (one

for each of the trials that are not part of any of S

, ..., S

r−1

, S

n−k+1

). Since there are



n−rk

r−1



permutations of r − 1 letters a and of n − rk − (r − 1) letters b, we have



<...<i

r−1

<n−k+1

P(E

···E

r−1

n−k+1

) =



n − rk

r − 1



(1 − p)

r−1

Putting it all together yields the exact expression, namely,

P(L

Ú k) =

n−k+1



r=1

(−1)

r+1





n − rk





n − rk

r − 1





(1 − p)

where we utilize the convention that





= 0ifm < j.

From a computational point of view, a more efﬁcient method for computing the

desired probability than the use of the preceding identity is to derive a set of recur-

sive equations. To do so, let A

be the event that there is a string of k consecutive

heads in a sequence of n ﬂips of a fair coin, and let P

= P(A

). We will derive a

set of recursive equations for P

by conditioning on when the ﬁrst tail appears. For

j = 1, ..., k, let F

be the event that the ﬁrst tail appears on ﬂip j, and let H be the

event that the ﬁrst k ﬂips are all heads. Because the events F

, ..., F

, H are mutually

exclusive and exhaustive (that is, exactly one of these events must occur), we have

P(A

) =



j=1

P(A

)P(F

) + P(A

|H)P(H)

Now, given that the ﬁrst tail appears on ﬂip j, where j < k, it follows that those j

ﬂips are wasted as far as obtaining a string of k heads in a row; thus, the conditional

probability of this event is the probability that such a string will occur among the

remaining n − j ﬂips. Therefore,

P(A

) = P

n−j

Because P(A

|H) = 1, the preceding equation gives

= P(A

)



j=1

n−j

P(F

) + P(H)



j=1

n−j

j−1

(1 − p) + p

Starting with P

= 0, j < k,andP

= p

, we can use the latter formula to recursively

compute P

k+1

, P

k+2

, and so on, up to P

. For instance, suppose we want the prob-

ability that there is a run of 2 consecutive heads when a fair coin is ﬂipped 4 times.

Then, with k = 2, we have P

= 0, P

= (1/2)

. Because, when p = 1/2, the recursion

becomes



j=1

n−j

(1/2)

+ (1/2)

152 Chapter 4 Random Variables

we obtain

= P

(1/2) + P

(1/2)

+ (1/2)

= 3/8

and

= P

(1/2) + P

(1/2)

+ (1/2)

= 1/2

which is clearly true because there are 8 outcomes that result in a string of 2 consec-

utive heads: hhhh, hhht, hhth, hthh, thhh, hhtt, thht,andtthh. Each of these outcomes

occurs with probability 1/16. .

Another use of the Poisson probability distribution arises in situations where

“events” occur at certain points in time. One example is to designate the occurrence

of an earthquake as an event; another possibility would be for events to correspond

to people entering a particular establishment (bank, post ofﬁce, gas station, and so

on); and a third possibility is for an event to occur whenever a war starts. Let us sup-

pose that events are indeed occurring at certain (random) points of time, and let us

assume that, for some positive constant λ, the following assumptions hold true:

1. The probability that exactly 1 event occurs in a given interval of length h is equal

to λh + o(h), where o(h) stands for any function f (h) for which lim

h→0

f (h)/h = 0.

[For instance, f (h) = h

is o(h), whereas f (h) = h is not.]

2. The probability that 2 or more events occur in an interval of length h is equal

to o(h).

3. For any integers n, j

, j

, ..., j

and any set of n nonoverlapping intervals, if we

deﬁne E

to be the event that exactly j

of the events under consideration occur

in the ith of these intervals, then events E

, E

, ..., E

are independent.

Loosely put, assumptions 1 and 2 state that, for small values of h, the probability

that exactly 1 event occurs in an interval of size h equals λh plus something that is

small compared with h, whereas the probability that 2 or more events occur is small

compared with h. Assumption 3 states that whatever occurs in one interval has no

(probability) effect on what will occur in other, nonoverlapping intervals.

We now show that, under assumptions 1, 2, and 3, the number of events occurring

in any interval of length t is a Poisson random variable with parameter λt.Tobe

precise, let us call the interval [0, t] and denote the number of events occurring in that

interval by N(t). To obtain an expression for P{N(t) = k}, we start by breaking the

interval [0, t]inton nonoverlapping subintervals, each of length t/n (Figure 4.7).

Now,

P{N(t) = k}=P{k of the n subintervals contain exactly 1 event

and the other n − k contain 0 events} (7.2)

+ P{N(t) = k and at least 1 subinterval contains

2 or more events}

The proceding equation holds because the event on the left side of Equation (7.2),

that is, {N(t) = k}, is clearly equal to the union of the two mutually exclusive events

–

—

–

(n – 1)

—

t =

FIGURE 4.7

Section 4.7 The Poisson Random Variable 153

on the right side of the equation. Letting A and B denote the two mutually exclusive

events on the right side of Equation (7.2), we have

P(B) … P{at least one subinterval contains 2 or more events}

= P

⎛

⎝



i=1

{ith subinterval contains 2 or more events}

⎞

⎠

…



i=1

P{ith subinterval contains 2 or more events}

by Boole’s

inequality



i=1





by assumption 2

= no





= t



o(t/n)

t/n



Now, in addition for any t, t/n

→0asn→q,soo(t /n)/(t/n)→0asn→q, by the deﬁni-

tion of o(h). Hence,

P(B)

→0asn→q (7.3)

Moreover, since assumptions 1 and 2 imply that

†

P{0 events occur in an interval of length h}

= 1 − [λh + o(h) + o(h)] = 1 − λh − o(h)

we see from the independence assumption (number 3) that

P(A) = P{k of the subintervals contain exactly 1 event and the other

n − k contain 0 events}







λt

+ o









1 −



λt



− o







n−k

However, since



λt

+ o







= λt + t



o(t/n)

t/n



→λt as n→q

it follows, by the same argument that veriﬁed the Poisson approximation to the bino-

mial, that

P(A)

→e

−λt

(λt)

as n

→q (7.4)

Thus, from Equations (7.2), (7.3), and (7.4), by letting n

→q, we obtain

P{N(t) = k}=e

−λt

(λt)

k = 0, 1, ... (7.5)

†

The sum of two functions, both of which are o(h), is also o(h). This is so because if

lim

h→0

f (h)/h = lim

h→0

g(h)/h = 0, then lim

h→0

[f (h) + g(h)]/h = 0.

154 Chapter 4 Random Variables

Hence, if assumptions 1, 2, and 3 are satisﬁed, then the number of events occurring

in any ﬁxed interval of length t is a Poisson random variable with mean λt, and we say

that the events occur in accordance with a Poisson process having rate λ. The value λ,

which can be shown to equal the rate per unit time at which events occur, is a constant

that must be empirically determined.

The preceding discussion explains why a Poisson random variable is usually a good

approximation for such diverse phenomena as the following:

1. The number of earthquakes occurring during some ﬁxed time span

2. The number of wars per year

3. The number of electrons emitted from a heated cathode during a ﬁxed time

period

4. The number of deaths, in a given period of time, of the policyholders of a life

insurance company

EXAMPLE 7e

Suppose that earthquakes occur in the western portion of the United States in accor-

dance with assumptions 1, 2, and 3, with λ = 2 and with 1 week as the unit of time.

(That is, earthquakes occur in accordance with the three assumptions at a rate of 2

per week.)

(a) Find the probability that at least 3 earthquakes occur during the next 2 weeks.

(b) Find the probability distribution of the time, starting from now, until the next

earthquake.

Solution. (a) From Equation (7.5), we have

P{N(2) Ú 3}=1 − P{N(2) = 0}−P{N(2) = 1}−P{N(2) = 2}

= 1 − e

−4

− 4e

−4

−

−4

= 1 − 13e

−4

(b) Let X denote the amount of time (in weeks) until the next earthquake. Because

X will be greater than t if and only if no events occur within the next t units of time,

we have, from Equation (7.5),

P{X > t}=P{N(t) = 0}=e

−λt

so the probability distribution function F of the random variable X is given by

F(t) = P{X … t}=1 − P{X > t}=1 − e

−λt

= 1 − e

−2t

4.7.1 Computing the Poisson Distribution Function

If X is Poisson with parameter λ,then

P{X = i + 1}

P{X = i}

−λ

i+1

/(i + 1)!

−λ

/i!

i + 1

(7.6)

Section 4.8 Other Discrete Probability Distributions 155

Starting with P{X = 0}=e

−λ

, we can use (7.6) to compute successively

P{X = 1}=λP{X = 0}

P{X = 2}=

P{X = 1}

P{X = i + 1}=

i + 1

P{X = i}

The website includes a program that uses Equation (7.6) to compute Poisson prob-

abilities.

EXAMPLE 7f

(a) Determine P{X … 90} when X is Poisson with mean 100.

(b) Determine P{Y … 1075} when Y is Poisson with mean 1000.

Solution. From the website, we obtain the solutions:

(a) P{X … 90} L .1714;

(b) P{Y … 1075} L .9894.

4.8 OTHER DISCRETE PROBABILITY DISTRIBUTIONS

4.8.1 The Geometric Random Variable

Suppose that independent trials, each having a probability p,0 < p < 1, of being a

success, are performed until a success occurs. If we let X equal the number of trials

required, then

P{X = n}=(1 − p)

n−1

pn= 1, 2, ... (8.1)

Equation (8.1) follows because, in order for X to equal n, it is necessary and sufﬁcient

that the ﬁrst n − 1 trials are failures and the nth trial is a success. Equation (8.1) then

follows, since the outcomes of the successive trials are assumed to be independent.

Since



n=1

P{X = n}=p



n=1

(1 − p)

n−1

1 − (1 − p)

= 1

it follows that, with probability 1, a success will eventually occur. Any random vari-

able X whose probability mass function is given by Equation (8.1) is said to be a

geometric random variable with parameter p.

EXAMPLE 8a

An urn contains N white and M black balls. Balls are randomly selected, one at a

time, until a black one is obtained. If we assume that each ball selected is replaced

before the next one is drawn, what is the probability that

(a) exactly n draws are needed?

(b) at least k draws are needed?

Solution. If we let X denote the number of draws needed to select a black ball, then

X satisﬁes Equation (8.1) with p = M/(M + N). Hence,

156 Chapter 4 Random Variables

(a)

P{X = n}=



M + N



n−1

M + N

n−1

(M + N)

(b)

P{X Ú k}=

M + N



n=k



M + N



n−1



M + N



M + N



k−1



1 −

M + N





M + N



k−1

Of course, part (b) could have been obtained directly, since the probability that at

least k trials are necessary to obtain a success is equal to the probability that the ﬁrst

k − 1 trials are all failures. That is, for a geometric random variable,

P{X Ú k}=(1 − p)

k−1

EXAMPLE 8b

Find the expected value of a geometric random variable.

Solution. With q = 1 − p, we have

E[X] =



i=1

i−1



i=1

(i − 1 + 1)q

i−1



i=1

(i − 1)q

i−1

p +



i=1

i−1



j=0

p + 1

= q



j=1

j−1

p + 1

= qE[X] + 1

Hence,

pE[X] = 1

yielding the result

E[X] =

Section 4.8 Other Discrete Probability Distributions 157

In other words, if independent trials having a common probability p of being success-

ful are performed until the ﬁrst success occurs, then the expected number of required

trials equals 1/p. For instance, the expected number of rolls of a fair die that it takes

to obtain the value 1 is 6. .

EXAMPLE 8c

Find the variance of a geometric random variable.

Solution. To determine Var(X), let us ﬁrst compute E[X

]. With q = 1 − p, we have

E[X

] =



i=1

i−1



i=1

(i − 1 + 1)

i−1



i=1

(i − 1)

i−1

p +



i=1

2(i − 1)q

i−1

p +



i=1

i−1



j=0

p + 2



j=1

p + 1

= qE[X

] + 2qE[X] + 1

Using E[X] = 1/p, the equation for E[X

] yields

pE[X

] =

+ 1

Hence,

E[X

] =

2q + p

q + 1

giving the result

Var(X) =

q + 1

−

1 − p

4.8.2 The Negative Binomial Random Variable

Suppose that independent trials, each having probability p,0 < p < 1, of being a

success are performed until a total of r successes is accumulated. If we let X equal the

number of trials required, then

P{X = n}=



n − 1

r − 1



(1 − p)

n−r

n = r, r + 1, ... (8.2)

Equation (8.2) follows because, in order for the rth success to occur at the nth trial,

there must be r − 1 successes in the ﬁrst n − 1 trials and the nth trial must be a

success. The probability of the ﬁrst event is



n − 1

r − 1



r−1

(1 − p)

n−r

158 Chapter 4 Random Variables

and the probability of the second is p; thus, by independence, Equation (8.2) is estab-

lished. To verify that a total of r successes must eventually be accumulated, either we

can prove analytically that



n=r

P{X = n}=



n=r



n − 1

r − 1



(1 − p)

n−r

= 1 (8.3)

or we can give a probabilistic argument as follows: The number of trials required

to obtain r successes can be expressed as Y

+ Y

+ ··· + Y

, where Y

equals

the number of trials required for the ﬁrst success, Y

the number of additional trials

after the ﬁrst success until the second success occurs, Y

the number of additional

trials until the third success, and so on. Because the trials are independent and all

have the same probability of success, it follows that Y

, Y

, ..., Y

are all geometric

random variables. Hence, each is ﬁnite with probability 1, so



i=1

must also be ﬁnite,

establishing Equation (8.3).

Any random variable X whose probability mass function is given by Equation (8.2)

is said to be a negative binomial random variable with parameters (r, p). Note that a

geometric random variable is just a negative binomial with parameter (1, p).

In the next example, we use the negative binomial to obtain another solution of

the problem of the points.

EXAMPLE 8d

If independent trials, each resulting in a success with probability p, are performed,

what is the probability of r successes occurring before m failures?

Solution. The solution will be arrived at by noting that r successes will occur before

m failures if and only if the rth success occurs no later than the (r + m − 1)th

trial. This follows because if the rth success occurs before or at the (r + m − 1)th

trial, then it must have occurred before the mth failure, and conversely. Hence, from

Equation (8.2), the desired probability is

r+m−1



n=r



n − 1

r − 1



(1 − p)

n−r

EXAMPLE 8e The Banach match problem

At all times, a pipe-smoking mathematician carries 2 matchboxes—1 in his left-hand

pocket and 1 in his right-hand pocket. Each time he needs a match, he is equally likely

to take it from either pocket. Consider the moment when the mathematician ﬁrst

discovers that one of his matchboxes is empty. If it is assumed that both matchboxes

initially contained N matches, what is the probability that there are exactly k matches,

k = 0, 1, ..., N, in the other box?

Solution. Let E denote the event that the mathematician ﬁrst discovers that the right-

hand matchbox is empty and that there are k matches in the left-hand box at the

time. Now, this event will occur if and only if the (N + 1)th choice of the right-hand

matchbox is made at the (N + 1 + N − k)th trial. Hence, from Equation (8.2) (with

p =

, r = N + 1, and n = 2N − k + 1), we see that

P(E) =



2N − k





2N−k+1

Section 4.8 Other Discrete Probability Distributions 159

Since there is an equal probability that it is the left-hand box that is ﬁrst discovered

to be empty and there are k matches in the right-hand box at that time, the desired

result is

2P(E) =



2N − k





2N−k

EXAMPLE 8f

Compute the expected value and the variance of a negative binomial random variable

with parameters r and p.

Solution. We have

E[X

] =



n=r



n − 1

r − 1



(1 − p)

n−r



n=r

k−1





r+1

(1 − p)

n−r

since n



n − 1

r − 1



= r







m=r+1

(m − 1)

k−1



m − 1



r+1

(1 − p)

m−(r+1)

by setting

m = n + 1

E[(Y − 1)

k−1

]

where Y is a negative binomial random variable with parameters r + 1, p. Setting

k = 1 in the preceding equation yields

E[X] =

Setting k = 2 in the equation for E[X

] and using the formula for the expected value

of a negative binomial random variable gives

E[X

] =

E[Y − 1]



r + 1

− 1



Therefore,

Var(X) =



r + 1

− 1



−





r(1 − p)

Thus, from Example 8f, if independent trials, each of which is a success with prob-

ability p, are performed, then the expected value and variance of the number of trials

that it takes to amass r successes is r/p and r(1 − p)/p

, respectively.

Since a geometric random variable is just a negative binomial with parameter

r = 1, it follows from the preceding example that the variance of a geometric ran-

dom variable with parameter p is equal to (1 − p)/p

, which checks with the result

of Example 8c.

160 Chapter 4 Random Variables

EXAMPLE 8g

Find the expected value and the variance of the number of times one must throw a

die until the outcome 1 has occurred 4 times.

Solution. Since the random variable of interest is a negative binomial with parame-

ters r = 4andp =

, it follows that

E[X] = 24

Var(X) =









= 120

4.8.3 The Hypergeometric Random Variable

Suppose that a sample of size n is to be chosen randomly (without replacement) from

an urn containing N balls, of which m are white and N − m are black. If we let X

denote the number of white balls selected, then

P{X = i}=





N − m

n − i







i = 0, 1, ..., n (8.4)

A random variable X whose probability mass function is given by Equation (8.4) for

some values of n, N, m is said to be a hypergeometric random variable.

Remark. Although we have written the hypergeometric probability mass func-

tion with i going from 0 to n, P{X = i} will actually be 0, unless i satisﬁes the inequali-

ties n − (N − m) … i … min(n, m). However, Equation (8.4) is always valid because

of our convention that





is equal to 0 when either k < 0orr < k. .

EXAMPLE 8h

An unknown number, say, N, of animals inhabit a certain region. To obtain some

information about the size of the population, ecologists often perform the follow-

ing experiment: They ﬁrst catch a number, say, m, of these animals, mark them in

some manner, and release them. After allowing the marked animals time to disperse

throughout the region, a new catch of size, say, n, is made. Let X denote the number

of marked animals in this second capture. If we assume that the population of ani-

mals in the region remained ﬁxed between the time of the two catches and that each

time an animal was caught it was equally likely to be any of the remaining uncaught

animals, it follows that X is a hypergeometric random variable such that

P{X = i}=





− m

n − i







K P

(N)

Suppose now that X is observed to equal i. Then, since P

(N) represents the proba-

bility of the observed event when there are actually N animals present in the region, it

would appear that a reasonable estimate of N would be the value of N that maximizes

(N). Such an estimate is called a maximum likelihood estimate. (See Theoretical

Exercises 13 and 18 for other examples of this type of estimation procedure.)

Section 4.8 Other Discrete Probability Distributions 161

The maximization of P

(N) can be done most simply by ﬁrst noting that

(N)

(N − 1)

(N − m)(N − n)

N(N − m − n + i)

Now, the preceding ratio is greater than 1 if and only if

(N − m)(N − n) Ú N(N − m − n + i)

or, equivalently, if and only if

N …

Thus, P

(N) is ﬁrst increasing and then decreasing, and reaches its maximum value

at the largest integral value not exceeding mn/i. This value is the maximum likeli-

hood estimate of N. For example, suppose that the initial catch consisted of m =

50 animals, which are marked and then released. If a subsequent catch consists of

n = 40 animals of which i = 4 are marked, then we would estimate that there are

some 500 animals in the region. (Note that the preceding estimate could also have

been obtained by assuming that the proportion of marked animals in the region,

m/N, is approximately equal to the proportion of marked animals in our second

catch, i/n.) .

EXAMPLE 8i

A purchaser of electrical components buys them in lots of size 10. It is his policy

to inspect 3 components randomly from a lot and to accept the lot only if all 3 are

nondefective. If 30 percent of the lots have 4 defective components and 70 percent

have only 1, what proportion of lots does the purchaser reject?

Solution. Let A denote the event that the purchaser accepts a lot. Now,

P(A) = P(A|lot has 4 defectives)

+ P(A|lot has 1 defective)





























100

Hence, 46 percent of the lots are rejected. .

If n balls are randomly chosen without replacement from a set of N balls of which

the fraction p = m/N is white, then the number of white balls selected is hypergeo-

metric. Now, it would seem that when m and N are large in relation to n, it shouldn’t

make much difference whether the selection is being done with or without replace-

ment, because, no matter which balls have previously been selected, when m and N

are large, each additional selection will be white with a probability approximately

equal to p. In other words, it seems intuitive that when m and N are large in relation

to n, the probability mass function of X should approximately be that of a binomial

random variable with parameters n and p. To verify this intuition, note that if X is

hypergeometric, then, for i … n,

162 Chapter 4 Random Variables

P{X = i}=





N − m

n − i







(m − i)! i!

(N − m)!

(N − m − n + i)! (n − i)!

(N − n)! n!





m − 1

N − 1

···

m − i + 1

N − i + 1

N − m

N − i

N − m − 1

N − i − 1

···

N − m − (n − i − 1)

N − i − (n − i − 1)





(1 − p)

n−i

when p = m/N and m and N are

large in relation to n and i

EXAMPLE 8j

Determine the expected value and the variance of X, a hypergeometric random vari-

able with parameters n, N,andm.

Solution.

E[X

] =



i=0

P{X = i}



i=1





N − m

n − i







Using the identities





= m



m − 1

i − 1



and n





= N



N − 1

n − 1



we obtain

E[X

] =



i=1

k−1



m − 1

i − 1



N − m

n − i





N − 1

n − 1



n−1



j=0

(j + 1)

k−1



m − 1



N − m

n − 1 − j





N − 1

n − 1



E[(Y + 1)

k−1

]

where Y is a hypergeometric random variable with parameters n − 1, N − 1, and

m − 1. Hence, upon setting k = 1, we have

E[X] =

In words, if n balls are randomly selected from a set of N balls, of which m are white,

then the expected number of white balls selected is nm/N.

Section 4.8 Other Discrete Probability Distributions 163

Upon setting k = 2 in the equation for E[X

], we obtain

E[X

] =

E[Y + 1]



(n − 1)(m − 1)

N − 1

+ 1



where the ﬁnal equality uses our preceding result to compute the expected value of

the hypergeometric random variable Y.

Because E[X] = nm/N, we can conclude that

Var(X) =



(n − 1)(m − 1)

N − 1

+ 1 −



Letting p = m/N and using the identity

m − 1

N − 1

Np − 1

N − 1

= p −

1 − p

N − 1

shows that

Var(X) = np[(n − 1)p − (n − 1)

1 − p

N − 1

+ 1 − np]

= np(1 − p)(1 −

n − 1

N − 1

)

Remark. We have shown in Example 8j that if n balls are randomly selected with-

out replacement from a set of N balls, of which the fraction p are white, then the

expected number of white balls chosen is np. In addition, if N is large in relation to n

[so that (N − n)/(N − 1) is approximately equal to 1], then

Var(X) L np(1 − p)

In other words, E[X] is the same as when the selection of the balls is done with

replacement (so that the number of white balls is binomial with parameters n

and p), and if the total collection of balls is large, then Var(X) is approximately equal

to what it would be if the selection were done with replacement. This is, of course,

exactly what we would have guessed, given our earlier result that when the number

of balls in the urn is large, the number of white balls chosen approximately has the

mass function of a binomial random variable. .

4.8.4 The Zeta (or Zipf) Distribution

A random variable is said to have a zeta (sometimes called the Zipf) distribution if

its probability mass function is given by

P{X = k}=

α+1

k = 1, 2, ...

for some value of α>0. Since the sum of the foregoing probabilities must equal 1, it

follows that

C =

⎡

⎣



k=1





α+1

⎤

⎦

−1

164 Chapter 4 Random Variables

The zeta distribution owes its name to the fact that the function

ζ(s) = 1 +









+ ··· +





+ ···

is known in mathematical disciplines as the Riemann zeta function (after the German

mathematician G. F. B. Riemann).

The zeta distribution was used by the Italian economist V. Pareto to describe the

distribution of family incomes in a given country. However, it was G. K. Zipf who

applied zeta distribution to a wide variety of problems in different areas and, in doing

so, popularized its use.

4.9 EXPECTED VALUE OF SUMS OF RANDOM VARIABLES

A very important property of expectations is that the expected value of a sum of

random variables is equal to the sum of their expectations. In this section, we will

prove this result under the assumption that the set of possible values of the proba-

bility experiment—that is, the sample space S—is either ﬁnite or countably inﬁnite.

Although the result is true without this assumption (and a proof is outlined in the

theoretical exercises), not only will the assumption simplify the argument, but it will

also result in an enlightening proof that will add to our intuition about expectations.

So, for the remainder of this section, suppose that the sample space S is either a ﬁnite

or a countably inﬁnite set.

For a random variable X, let X(s) denote the value of X when s ∈ S is the outcome

of the experiment. Now, if X and Y are both random variables, then so is their sum.

That is, Z = X + Y is also a random variable. Moreover, Z(s) = X(s) + Y(s).

EXAMPLE 9a

Suppose that the experiment consists of ﬂipping a coin 5 times, with the outcome

being the resulting sequence of heads and tails. Suppose X is the number of heads in

theﬁrst3ﬂipsandY is the number of heads in the ﬁnal 2 ﬂips. Let Z = X + Y. Then,

for instance, for the outcome s = (h, t, h, t , h),

X(s) = 2

Y(s) = 1

Z(s) = X(s) + Y(s) = 3

meaning that the outcome (h, t, h, t, h) results in 2 heads in the ﬁrst three ﬂips, 1 head

in the ﬁnal two ﬂips, and a total of 3 heads in the ﬁve ﬂips. .

Let p(s) = P({s}) be the probability that s is the outcome of the experiment.

Because we can write any event A as the ﬁnite or countably inﬁnite union of the

mutually exclusive events {s}, s ∈ A, it follows by the axioms of probability that

P(A) =



s∈A

p(s)

When A = S, the preceding equation gives

1 =



s∈S

p(s)

Section 4.9 Expected Value of Sums of Random Variables 165

Now, let X be a random variable, and consider E[X]. Because X(s) is the value of X

when s is the outcome of the experiment, it seems intuitive that E[X]—the weighted

average of the possible values of X, with each value weighted by the probability that

X assumes that value—should equal a weighted average of the values X(s), s ∈ S,

with X(s) weighted by the probability that s is the outcome of the experiment. We

now prove this intuition.

Proposition 9.1.

E[X] =



s∈S

X(s) p(s)

Proof. Suppose that the distinct values of X are x

, i Ú 1. For each i, let S

be the

event that X is equal to x

. That is, S

={s : X(s) = x

}. Then,

E[X] =



P{X = x

}



P(S

)



s∈S

p(s)



s∈S

p(s)



s∈S

X(s)p(s)



s∈S

X(s)p(s)

where the ﬁnal equality follows because S

, S

, ... are mutually exclusive events

whose union is S.

EXAMPLE 9b

Suppose that two independent ﬂips of a coin that comes up heads with probability p

are made, and let X denote the number of heads obtained. Because

P(X = 0) = P(t, t) = (1 − p)

P(X = 1) = P(h, t) + P(t, h) = 2p(1 − p)

P(X = 2) = P(h, h) = p

it follows from the deﬁnition of expected value that

E[X] = 0 · (1 − p)

+ 1 · 2p(1 − p) + 2 · p

= 2p

which agrees with

E[X] = X(h, h)p

+ X(h, t)p(1 − p) + X(t, h)(1 − p)p + X(t, t)(1 − p)

= 2p

+ p(1 − p) + (1 − p)p

= 2p .

We now prove the important and useful result that the expected value of a sum of

random variables is equal to the sum of their expectations.

166 Chapter 4 Random Variables

Corollary 9.2. For random variables X

, X

, ..., X

⎡

⎣



i=1

⎤

⎦



i=1

E[X

]

Proof. Let Z =



i=1

. Then, by Proposition 9.1,

E[Z] =



s∈S

Z(s)p(s)



s∈S



(s) + X

(s) + ... + X

(s)



p(s)



s∈S

(s)p(s) +



s∈S

(s)p(s) + ... +



s∈S

(s)p(s)

= E[X

] + E[X

] + ... + E[X

]

EXAMPLE 9c

Find the expected value of the sum obtained when n fair dice are rolled.

Solution. Let X be the sum. We will compute E [X] by using the representation

X =



i=1

where X

is the upturned value on die i. Because X

is equally likely to be any of the

values from 1 to 6, it follows that

E[X

] =



i=1

i(1/6) = 21/6 = 7/2

which yields the result

E[X] = E

⎡

⎣



i=1

⎤

⎦



i=1

E[X

] = 3.5 n

EXAMPLE 9d

Find the expected total number of successes that result from n trials when trial i is a

success with probability p

, i = 1, ..., n.

Solution. Letting

1, if trial i is a success

0, if trial i is a failure

we have the representation

X =



i=1

Section 4.9 Expected Value of Sums of Random Variables 167

Consequently,

E[X] =



i=1

E[X

] =



i=1

Note that this result does not require that the trials be independent. It includes as a

special case the expected value of a binomial random variable, which assumes inde-

pendent trials and all p

= p, and thus has mean np. It also gives the expected value

of a hypergeometric random variable representing the number of white balls selected

when n balls are randomly selected, without replacement, from an urn of N balls of

which m are white. We can interpret the hypergeometric as representing the number

of successes in n trials, where trial i is said to be a success if the ith ball selected is

white. Because the ith ball selected is equally likely to be any of the N balls and thus

has probability m/N of being white, it follows that the hypergeometric is the num-

ber of successes in n trials in which each trial is a success with probability p = m/N.

Hence, even though these hypergeometric trials are dependent, it follows from the

result of Example 9d that the expected value of the hypergeometric is np = nm/N. .

EXAMPLE 9e

Derive an expression for the variance of the number of successful trials in Example

9d, and apply it to obtain the variance of a binomial random variable with parameters

n and p, and of a hypergeometric random variable equal to the number of white balls

chosen when n balls are randomly chosen from an urn containing N balls of which m

are white.

Solution. Letting X be the number of successful trials, and using the same represen-

tation for X—namely, X =



i=1

—as in the previous example, we have

E[X

] = E

⎡

⎢

⎣

⎛

⎝



i=1

⎞

⎠

⎛

⎜

⎝



j=1

⎞

⎟

⎠

⎤

⎥

⎦

= E

⎡

⎢

⎣



i=1

⎛

⎝



jZi

⎞

⎠

⎤

⎥

⎦

= E

⎡

⎣



i=1



i=1



jZi

⎤

⎦



i=1

E[X

] +



i=1



jZi

E[X

]



i=1



jZi

E[X

] (9.1)

where the ﬁnal equation used that X

= X

. However, because the possible values of

both X

and X

are 0 or 1, it follows that

1, if X

= 1, X

= 1

0, otherwise

168 Chapter 4 Random Variables

Hence,

E[X

] = P{X

= 1, X

= 1}=P(trials i and j are successes)

Now, on the one hand, if X is binomial, then, for i Z j, the results of trial i and trial j

are independent, with each being a success with probability p. Therefore,

E[X

] = p

, i Z j

Together with Equation (9.1), the preceding equation shows that, for a binomial ran-

dom variable X,

E[X

] = np + n(n − 1)p

implying that

Var(X) = E[X

] − (E[X])

= np + n(n − 1)p

− n

= np(1 − p)

On the other hand, if X is hypergeometric, then, given that a white ball is chosen

in trial i, each of the other N − 1 balls, of which m − 1 are white, is equally likely to

be the jth ball chosen, for j Z i. Consequently, for j Z i,

P{X

= 1, X

= 1}=P{X

= 1}P{X

= 1|X

= 1}=

m − 1

N − 1

Using p

= m/N, we now obtain, from Equation (9.1),

E[X

] =

+ n(n − 1)

m − 1

N − 1

Consequently,

Var(X) =

+ n(n − 1)

m − 1

N − 1

−





which, as shown in Example 8j, can be simpliﬁed to yield

Var(X) = np(1 − p)



1 −

n − 1

N − 1



where p = m/N. .

4.10 PROPERTIES OF THE CUMULATIVE DISTRIBUTION FUNCTION

Recall that, for the distribution function F of X, F(b) denotes the probability that the

random variable X takes on a value that is less than or equal to b. Following are some

properties of the cumulative distribution function (c.d.f.) F:

1. F is a nondecreasing function; that is, if a < b,thenF(a) … F(b).

2. lim

b→q

F(b) = 1.

3. lim

b→−q

F(b) = 0.

4. F is right continuous. That is, for any b and any decreasing sequence b

, n Ú 1,

that converges to b, lim

n→q

F(b

) = F(b).

Property 1 follows, as was noted in Section 4.1, because, for a < b, the event

{X … a} is contained in the event {X … b} and so cannot have a larger probabil-

ity. Properties 2, 3, and 4 all follow from the continuity property of probabilities

Section 4.10 Properties of the Cumulative Distribution Function 169

(Section 2.6). For instance, to prove property 2, we note that if b

increases to q,then

the events {X … b

}, n Ú 1, are increasing events whose union is the event {X < q}.

Hence, by the continuity property of probabilities,

lim

n→q

P{X … b

}=P{X < q}=1

which proves property 2.

The proof of property 3 is similar and is left as an exercise. To prove property 4,

we note that if b

decreases to b,then{X … b

}, n Ú 1, are decreasing events whose

intersection is {X … b}. The continuity property then, yields

lim

P{X … b

}=P{X … b}

which veriﬁes property 4.

All probability questions about X can be answered in terms of the c.d.f., F.For

example,

P{a < X … b}=F(b) − F(a) for all a < b (8.1)

This equation can best be seen to hold if we write the event {X … b} as the union of

the mutually exclusive events {X … a} and {a < X … b}. That is,

{X … b}={X … a}∪{a < X … b}

P{X … b}=P{X … a}+P{a < X … b}

which establishes Equation (9.1).

If we want to compute the probability that X is strictly less than b, we can again

apply the continuity property to obtain

P{X < b}=P



lim

n→q

X … b −



= lim

n→q



X … b −



= lim

n→q



b −



Note that P{X < b} does not necessarily equal F(b), since F(b) also includes the

probability that X equals b.

EXAMPLE 10a

The distribution function of the random variable X is given by

F(x) =

⎧

⎪

⎨

⎪

⎩

0 x < 0

0 … x < 1

1 … x < 2

2 … x < 3

13… x

170 Chapter 4 Random Variables

–

—

123

(

)

FIGURE 4.8: Graph of F(x).

A graph of F(x) is presented in Figure 4.8. Compute (a) P {X < 3},(b)P{X = 1}, (c)

P{X >

},and(d)P{2 < X … 4}.

Solution. (a) P{X < 3}=lim

X … 3 −

= lim



3 −



(b)

P{X = 1}=P{X … 1}−P{X < 1}

= F(1) − lim



1 −



−

(c)

X >

= 1 − P

X …

= 1 − F





(d)

P{2 < X … 4}=F(4) − F(2)

SUMMARY

A real-valued function deﬁned on the outcome of a probability experiment is called

a random variable.

If X is a random variable, then the function F(x) deﬁned by

F(x) = P{X … x}

is called the distribution function of X. All probabilities concerning X can be stated

in terms of F.

Summary 171

A random variable whose set of possible values is either ﬁnite or countably inﬁnite

is called discrete.IfX is a discrete random variable, then the function

p(x) = P{X = x}

is called the probability mass function of X. Also, the quantity E[X] deﬁned by

E[X] =



x:p(x)>0

xp(x)

is called the expected value of X. E[X] is also commonly called the mean or the expec-

tation of X.

A useful identity states that, for a function g,

E[g(X)] =



x:p(x)>0

g(x)p(x)

The variance of a random variable X, denoted by Var(X), is deﬁned by

Var(X) = E[(X − E[X])

]

The variance, which is equal to the expected square of the difference between X

and its expected value, is a measure of the spread of the possible values of X. A useful

identity is

Var(X) = E[X

] − (E[X])

The quantity

√

Var(X) is called the standard deviation of X.

We now note some common types of discrete random variables. The random vari-

able X whose probability mass function is given by

p(i) =





(1 − p)

n−i

i = 0, ..., n

is said to be a binomial random variable with parameters n and p. Such a random

variable can be interpreted as being the number of successes that occur when n inde-

pendent trials, each of which results in a success with probability p, are performed.

Its mean and variance are given by

E[X] = np Var(X) = np(1 − p)

The random variable X whose probability mass function is given by

p(i) =

−λ

i Ú 0

is said to be a Poisson random variable with parameter λ. If a large number of (approx-

imately) independent trials are performed, each having a small probability of being

successful, then the number of successful trials that result will have a distribution

which is approximately that of a Poisson random variable. The mean and variance of

a Poisson random variable are both equal to its parameter λ. That is,

E[X] = Var(X) = λ

The random variable X whose probability mass function is given by

p(i) = p(1 − p)

i−1

i = 1, 2, ...

172 Chapter 4 Random Variables

is said to be a geometric random variable with parameter p. Such a random variable

represents the trial number of the ﬁrst success when each trial is independently a

success with probability p. Its mean and variance are given by

E[X] =

Var(X) =

1 − p

The random variable X whose probability mass function is given by

p(i) =



i − 1

r − 1



(1 − p)

i−r

i Ú r

is said to be a negative binomial random variable with parameters r and p. Such a

random variable represents the trial number of the rth success when each trial is

independently a success with probability p. Its mean and variance are given by

E[X] =

Var(X) =

r(1 − p)

A hypergeometric random variable X with parameters n, N,andm represents the

number of white balls selected when n balls are randomly chosen from an urn that

contains N balls of which m are white. The probability mass function of this random

variable is given by

p(i) =





N − m

n − i







i = 0, ..., m

With p = m/N, its mean and variance are

E[X] = np Var(X) =

N − n

N − 1

np(1 − p)

An important property of the expected value is that the expected value of a sum of

random variables is equal to the sum of their expected values. That is,

⎡

⎣



i=1

⎤

⎦



i=1

E[X

]

PROBLEMS

4.1. Two balls are chosen randomly from an urn con-

taining 8 white, 4 black, and 2 orange balls. Sup-

pose that we win $2 for each black ball selected

and we lose $1 for each white ball selected. Let

X denote our winnings. What are the possible val-

ues of X, and what are the probabilities associated

with each value?

4.2. Two fair dice are rolled. Let X equal the

product of the 2 dice. Compute P{X = i} for

i = 1, ..., 36.

4.3. Three dice are rolled. By assuming that each of

the 6

= 216 possible outcomes is equally likely,

ﬁnd the probabilities attached to the possible val-

ues that X can take on, where X is the sum of

the 3 dice.

4.4. Five men and 5 women are ranked according to

their scores on an examination. Assume that no

two scores are alike and all 10! possible rankings

are equally likely. Let X denote the highest rank-

ing achieved by a woman. (For instance, X = 1

Problems 173

if the top-ranked person is female.) Find P{X=i},

i = 1, 2, 3, ...,8,9,10.

4.5. Let X represent the difference between the num-

ber of heads and the number of tails obtained

when a coin is tossed n times. What are the pos-

sible values of X?

4.6. In Problem 5, for n = 3, if the coin is assumed fair,

what are the probabilities associated with the val-

ues that X can take on?

4.7. Suppose that a die is rolled twice. What are the

possible values that the following random vari-

ables can take on:

(a) the maximum value to appear in the two rolls;

(b) the minimum value to appear in the two rolls;

(d) the value of the ﬁrst roll minus the value of

the second roll?

4.8. If the die in Problem 7 is assumed fair, calculate

the probabilities associated with the random vari-

ables in parts (a) through (d).

4.9. Repeat Example 1b when the balls are selected

with replacement.

4.10. In Example 1d, compute the conditional probabil-

ity that we win i dollars, given that we win some-

thing; compute it for i = 1, 2, 3.

4.11. (a) An integer N is to be selected at random from

{1, 2, ..., (10)

} in the sense that each integer

has the same probability of being selected.

What is the probability that N will be divis-

ible by 3? by 5? by 7? by 15? by 105? How

would your answer change if (10)

is replaced

by (10)

as k became larger and larger?

(b) An important function in number theory—

one whose properties can be shown to be

related to what is probably the most impor-

tant unsolved problem of mathematics, the

Riemann hypothesis—is the M

obius function

μ(n), deﬁned for all positive integral values n

as follows: Factor n into its prime factors. If

there is a repeated prime factor, as in 12 =

2 · 2 · 3or49= 7 · 7, then μ(n) is deﬁned

to equal 0. Now let N be chosen at random

from {1, 2, ...(10)

},wherek is large. Deter-

mine P{μ(N) = 0} as k

→q.

Hint: To compute P{μ(N) Z 0}, use the identity



i=1

− 1







···=

where P

is the ith-smallest prime. (The number 1

is not a prime.)

4.12. In the game of Two-Finger Morra, 2 players show

1 or 2 ﬁngers and simultaneously guess the number

of ﬁngers their opponent will show. If only one of

the players guesses correctly, he wins an amount

(in dollars) equal to the sum of the ﬁngers shown

by him and his opponent. If both players guess

correctly or if neither guesses correctly, then no

money is exchanged. Consider a speciﬁed player,

and denote by X the amount of money he wins in

a single game of Two-Finger Morra.

(a) If each player acts independently of the other,

and if each player makes his choice of the

number of ﬁngers he will hold up and the

number he will guess that his opponent will

hold up in such a way that each of the 4 pos-

sibilities is equally likely, what are the possi-

ble values of X and what are their associated

probabilities?

(b) Suppose that each player acts independently

of the other. If each player decides to hold up

the same number of ﬁngers that he guesses his

opponent will hold up, and if each player is

equally likely to hold up 1 or 2 ﬁngers, what

are the possible values of X and their associ-

ated probabilities?

4.13. A salesman has scheduled two appointments to

sell encyclopedias. His ﬁrst appointment will lead

to a sale with probability .3, and his second will

lead independently to a sale with probability .6.

Any sale made is equally likely to be either for the

deluxe model, which costs $1000, or the standard

model, which costs $500. Determine the probabil-

ity mass function of X, the total dollar value of all

sales.

4.14. Five distinct numbers are randomly distributed

to players numbered 1 through 5. Whenever two

players compare their numbers, the one with the

higher one is declared the winner. Initially, players

1 and 2 compare their numbers; the winner then

compares her number with that of player 3, and so

on. Let X denote the number of times player 1 is a

winner. Find P{X = i}, i = 0, 1, 2, 3, 4.

4.15. The National Basketball Association (NBA) draft

lottery involves the 11 teams that had the worst

won–lost records during the year. A total of 66

balls are placed in an urn. Each of these balls is

inscribed with the name of a team: Eleven have

the name of the team with the worst record, 10

have the name of the team with the second-worst

record, 9 have the name of the team with the third-

worst record, and so on (with 1 ball having the

name of the team with the 11th-worst record).

A ball is then chosen at random, and the team

whose name is on the ball is given the ﬁrst pick

in the draft of players about to enter the league.

Another ball is then chosen, and if it “belongs”

to a team different from the one that received the

ﬁrst draft pick, then the team to which it belongs

receives the second draft pick. (If the ball belongs

174 Chapter 4 Random Variables

to the team receiving the ﬁrst pick, then it is dis-

carded and another one is chosen; this continues

until the ball of another team is chosen.) Finally,

another ball is chosen, and the team named on

the ball (provided that it is different from the

previous two teams) receives the third draft pick.

The remaining draft picks 4 through 11 are then

awarded to the 8 teams that did not “win the lot-

tery,” in inverse order of their won–lost records.

For instance, if the team with the worst record did

not receive any of the 3 lottery picks, then that

team would receive the fourth draft pick. Let X

denote the draft pick of the team with the worst

record. Find the probability mass function of X.

4.16. In Problem 15, let team number 1 be the team

with the worst record, let team number 2 be the

team with the second-worst record, and so on. Let

denote the team that gets draft pick number i.

(Thus, Y

= 3 if the ﬁrst ball chosen belongs to

team number 3.) Find the probability mass func-

tion of (a) Y

,(b)Y

,and(c)Y

4.17. Suppose that the distribution function of X is

given by

F(b) =

⎧

⎪

⎨

⎪

⎩

0 b < 0

0 … b < 1

b − 1

1 … b < 2

2 … b < 3

13… b

(a) Find P{X = i}, i = 1, 2, 3.

(b) Find P{

< X <

4.18. Four independent ﬂips of a fair coin are made. Let

X denote the number of heads obtained. Plot the

probability mass function of the random variable

X − 2.

4.19. If the distribution function of X is given by

F(b) =

⎧

⎪

⎨

⎪

⎩

0 b < 0

0 … b < 1

1 … b < 2

2 … b < 3

3 … b < 3.5

1 b Ú 3.5

calculate the probability mass function of X.

4.20. A gambling book recommends the following “win-

ning strategy” for the game of roulette: Bet $1 on

red. If red appears



which has probability



,then

take the $1 proﬁt and quit. If red does not appear

and you lose this bet



which has probability

occurring



, make additional $1 bets on red on each

of the next two spins of the roulette wheel and then

quit. Let X denote your winnings when you quit.

(a) Find P{X > 0}.

(b) Are you convinced that the strategy is indeed

a “winning” strategy? Explain your answer!

4.21. Four buses carrying 148 students from the same

school arrive at a football stadium. The buses

carry, respectively, 40, 33, 25, and 50 students. One

of the students is randomly selected. Let X denote

the number of students that were on the bus car-

rying the randomly selected student. One of the 4

bus drivers is also randomly selected. Let Y denote

the number of students on her bus.

(a) Which of E[X]orE[Y] do you think is larger?

Why?

(b) Compute E[X]andE[Y].

4.22. Suppose that two teams play a series of games that

ends when one of them has won i games. Suppose

that each game played is, independently, won by

team A with probability p. Find the expected num-

ber of games that are played when (a) i = 2and(b)

i = 3. Also, show in both cases that this number is

maximized when p =

4.23. You have $1000, and a certain commodity

presently sells for $2 per ounce. Suppose that after

one week the commodity will sell for either $1

or $4 an ounce, with these two possibilities being

equally likely.

(a) If your objective is to maximize the expected

amount of money that you possess at the

end of the week, what strategy should you

employ?

(b) If your objective is to maximize the expected

amount of the commodity that you possess at

the end of the week, what strategy should you

employ?

4.24. A and B play the following game: A writes down

either number 1 or number 2, and B must guess

which one. If the number that A has written down

is i and B has guessed correctly, B receives i units

from A.IfB makes a wrong guess, B pays

unit to

A.IfB randomizes his decision by guessing 1 with

probability p and 2 with probability 1 − p, deter-

mine his expected gain if (a) A has written down

number 1 and (b) A has written down number 2.

What value of p maximizes the minimum pos-

sible value of B’s expected gain, and what is

this maximin value? (Note that B’s expected

Problems 175

gain depends not only on p, but also on what

A does.)

Consider now player A. Suppose that she also

randomizes her decision, writing down number 1

with probability q. What is A’s expected loss if (c)

B chooses number 1 and (d) B chooses number 2?

What value of q minimizes A’s maximum

expected loss? Show that the minimum of A’s max-

imum expected loss is equal to the maximum of B’s

minimum expected gain. This result, known as the

minimax theorem, was ﬁrst established in general-

ity by the mathematician John von Neumann and

is the fundamental result in the mathematical disci-

pline known as the theory of games. The common

value is called the value of the game to player B.

4.25. Two coins are to be ﬂipped. The ﬁrst coin will land

on heads with probability .6, the second with prob-

ability .7. Assume that the results of the ﬂips are

independent, and let X equal the total number of

heads that result.

(a) Find P{X = 1}.

(b) Determine E[X].

4.26. One of the numbers 1 through 10 is randomly cho-

sen. You are to try to guess the number chosen by

asking questions with “yes–no” answers. Compute

the expected number of questions you will need to

ask in each of the following two cases:

(a) Your ithquestionistobe“Isiti?” i =

1, 2, 3, 4, 5, 6, 7, 8, 9, 10.

(b) With each question you try to eliminate one-

half of the remaining numbers, as nearly as

possible.

4.27. An insurance company writes a policy to the effect

that an amount of money A must be paid if some

event E occurs within a year. If the company esti-

mates that E will occur within a year with probabil-

ity p, what should it charge the customer in order

that its expected proﬁt will be 10 percent of A?

4.28. A sample of 3 items is selected at random from a

box containing 20 items of which 4 are defective.

Find the expected number of defective items in the

sample.

4.29. There are two possible causes for a breakdown of

a machine. To check the ﬁrst possibility would cost

dollars, and, if that were the cause of the break-

down, the trouble could be repaired at a cost of R

dollars. Similarly, there are costs C

and R

asso-

ciated with the second possibility. Let p and 1 −

p denote, respectively, the probabilities that the

breakdown is caused by the ﬁrst and second possi-

bilities. Under what conditions on p, C

, R

, i = 1, 2,

should we check the ﬁrst possible cause of break-

down and then the second, as opposed to reversing

the checking order, so as to minimize the expected

cost involved in returning the machine to working

order?

Note: If the ﬁrst check is negative, we must still

check the other possibility.

4.30. A person tosses a fair coin until a tail appears for

the ﬁrst time. If the tail appears on the nth ﬂip, the

person wins 2

dollars. Let X denote the player’s

winnings. Show that E[X] =+q. This problem is

known as the St. Petersburg paradox.

(a) Would you be willing to pay $1 million to play

this game once?

(b) Would you be willing to pay $1 million for

each game if you could play for as long as

you liked and only had to settle up when you

stopped playing?

4.31. Each night different meteorologists give us the

probability that it will rain the next day. To judge

how well these people predict, we will score each

of them as follows: If a meteorologist says that it

will rain with probability p, then he or she will

receive a score of

1 − (1 − p)

if it does rain

1 − p

if it does not rain

We will then keep track of scores over a cer-

tain time span and conclude that the meteorologist

with the highest average score is the best predictor

of weather. Suppose now that a given meteorolo-

gist is aware of our scoring mechanism and wants

to maximize his or her expected score. If this per-

son truly believes that it will rain tomorrow with

probability p

∗

, what value of p should he or she

assert so as to maximize the expected score?

4.32. To determine whether they have a certain dis-

ease, 100 people are to have their blood tested.

However, rather than testing each individual sepa-

rately, it has been decided ﬁrst to place the peo-

ple into groups of 10. The blood samples of the

10 people in each group will be pooled and ana-

lyzed together. If the test is negative, one test will

sufﬁce for the 10 people, whereas if the test is posi-

tive, each of the 10 people will also be individually

tested and, in all, 11 tests will be made on this

group. Assume that the probability that a person

has the disease is .1 for all people, independently

of each other, and compute the expected number

of tests necessary for each group. (Note that we are

assuming that the pooled test will be positive if at

least one person in the pool has the disease.)

4.33. A newsboy purchases papers at 10 cents and sells

them at 15 cents. However, he is not allowed to

return unsold papers. If his daily demand is a bino-

mial random variable with n = 10, p =

, approxi-

mately how many papers should he purchase so as

to maximize his expected proﬁt?

4.34. In Example 4b, suppose that the department store

incurs an additional cost of c for each unit of unmet

176 Chapter 4 Random Variables

demand. (This type of cost is often referred to as

a goodwill cost because the store loses the good-

will of those customers whose demands it can-

not meet.) Compute the expected proﬁt when the

store stocks s units, and determine the value of s

that maximizes the expected proﬁt.

4.35. A box contains 5 red and 5 blue marbles. Two mar-

bles are withdrawn randomly. If they are the same

color, then you win $1.10; if they are different col-

ors, then you win −$1.00. (That is, you lose $1.00.)

Calculate

(a) the expected value of the amount you win;

(b) the variance of the amount you win.

4.36. Consider Problem 22 with i = 2. Find the variance

of the number of games played, and show that this

number is maximized when p =

4.37. Find Var(X) and Var(Y) for X and Y as given in

Problem 21.

4.38. If E[X] = 1andVar(X) = 5, ﬁnd

(a) E[(2 + X)

];

(b) Var(4 + 3X).

4.39. A ball is drawn from an urn containing 3 white and

3 black balls. After the ball is drawn, it is replaced

and another ball is drawn. This process goes on

indeﬁnitely. What is the probability that, of the

ﬁrst 4 balls drawn, exactly 2 are white?

4.40. On a multiple-choice exam with 3 possible answers

for each of the 5 questions, what is the probability

that a student will get 4 or more correct answers

just by guessing?

4.41. A man claims to have extrasensory perception. As

a test, a fair coin is ﬂipped 10 times and the man is

asked to predict the outcome in advance. He gets

7 out of 10 correct. What is the probability that

he would have done at least this well if he had no

ESP?

4.42. Suppose that, in ﬂight, airplane engines will fail

with probability 1 − p, independently from engine

to engine. If an airplane needs a majority of its

engines operative to complete a successful ﬂight,

for what values of p is a 5-engine plane preferable

to a 3-engine plane?

4.43. A communications channel transmits the digits 0

and 1. However, due to static, the digit transmitted

is incorrectly received with probability .2. Suppose

that we want to transmit an important message

consisting of one binary digit. To reduce the

chance of error, we transmit 00000 instead of 0 and

11111 instead of 1. If the receiver of the message

uses “majority” decoding, what is the probability

that the message will be wrong when decoded?

What independence assumptions are you making?

4.44. A satellite system consists of n components and

functions on any given day if at least k of the n

components function on that day. On a rainy day

each of the components independently functions

with probability p

, whereas on a dry day they each

independently function with probability p

.Ifthe

probability of rain tomorrow is α, what is the prob-

ability that the satellite system will function?

4.45. A student is getting ready to take an important

oral examination and is concerned about the pos-

sibility of having an “on” day or an “off” day. He

ﬁgures that if he has an on day, then each of his

examiners will pass him, independently of each

other, with probability .8, whereas if he has an off

day, this probability will be reduced to .4. Sup-

pose that the student will pass the examination if a

majority of the examiners pass him. If the student

feels that he is twice as likely to have an off day as

he is to have an on day, should he request an exam-

ination with 3 examiners or with 5 examiners?

4.46. Suppose that it takes at least 9 votes from a 12-

member jury to convict a defendant. Suppose also

that the probability that a juror votes a guilty per-

son innocent is .2, whereas the probability that the

juror votes an innocent person guilty is .1. If each

juror acts independently and if 65 percent of the

defendants are guilty, ﬁnd the probability that the

jury renders a correct decision. What percentage

of defendants is convicted?

4.47. In some military courts, 9 judges are appointed.

However, both the prosecution and the defense

attorneys are entitled to a peremptory challenge

of any judge, in which case that judge is removed

from the case and is not replaced. A defendant is

declared guilty if the majority of judges cast votes

of guilty, and he or she is declared innocent other-

wise. Suppose that when the defendant is, in fact,

guilty, each judge will (independently) vote guilty

with probability .7, whereas when the defendant is,

in fact, innocent, this probability drops to .3.

(a) What is the probability that a guilty defendant

is declared guilty when there are (i) 9, (ii) 8,

and (iii) 7 judges?

(b) Repeat part (a) for an innocent defendant.

the right to a peremptory challenge of a judge,

and if the defense is limited to at most two

such challenges, how many challenges should

the defense attorney make if he or she is 60

percent certain that the client is guilty?

4.48. It is known that diskettes produced by a cer-

tain company will be defective with probability

.01, independently of each other. The company

sells the diskettes in packages of size 10 and

offers a money-back guarantee that at most 1 of

the 10 diskettes in the package will be defective.

The guarantee is that the customer can return the

entire package of diskettes if he or she ﬁnds more

Problems 177

than one defective diskette in it. If someone buys

3 packages, what is the probability that he or she

will return exactly 1 of them?

4.49. When coin 1 is ﬂipped, it lands on heads with prob-

ability .4; when coin 2 is ﬂipped, it lands on heads

with probability .7. One of these coins is randomly

chosen and ﬂipped 10 times.

(a) What is the probability that the coin lands on

heads on exactly 7 of the 10 ﬂips?

(b) Given that the ﬁrst of these ten ﬂips lands

heads, what is the conditional probability that

exactly 7 of the 10 ﬂips land on heads?

4.50. Suppose that a biased coin that lands on heads with

probability p is ﬂipped 10 times. Given that a total

of 6 heads results, ﬁnd the conditional probability

that the ﬁrst 3 outcomes are

(a) h, t, t (meaning that the ﬁrst ﬂip results in

heads, the second in tails, and the third in

tails);

(b) t, h, t.

4.51. The expected number of typographical errors on a

page of a certain magazine is .2. What is the prob-

ability that the next page you read contains (a) 0

and (b) 2 or more typographical errors? Explain

your reasoning!

4.52. The monthly worldwide average number of air-

plane crashes of commercial airlines is 3.5. What

is the probability that there will be

(a) at least 2 such accidents in the next month;

(b) at most 1 accident in the next month?

Explain your reasoning!

4.53. Approximately 80,000 marriages took place in the

state of New York last year. Estimate the proba-

bility that, for at least one of these couples,

(a) both partners were born on April 30;

(b) both partners celebrated their birthday on the

same day of the year.

State your assumptions.

4.54. Suppose that the average number of cars aban-

doned weekly on a certain highway is 2.2. Approx-

imate the probability that there will be

(a) no abandoned cars in the next week;

(b) at least 2 abandoned cars in the next week.

4.55. A certain typing agency employs 2 typists. The

average number of errors per article is 3 when

typed by the ﬁrst typist and 4.2 when typed by the

second. If your article is equally likely to be typed

by either typist, approximate the probability that it

will have no errors.

4.56. How many people are needed so that the probabil-

ity that at least one of them has the same birthday

as you is greater than

4.57. Suppose that the number of accidents occurring on

a highway each day is a Poisson random variable

with parameter λ = 3.

(a) Find the probability that 3 or more accidents

occur today.

(b) Repeat part (a) under the assumption that at

least 1 accident occurs today.

4.58. Compare the Poisson approximation with the cor-

rect binomial probability for the following cases:

(a) P{X = 2} when n = 8, p = .1;

(b) P{X = 9} when n = 10, p = .95;

(d) P{X = 4} when n = 9, p = .2.

4.59. If you buy a lottery ticket in 50 lotteries, in each of

which your chance of winning a prize is

100

,what

is the (approximate) probability that you will win

aprize

(a) at least once?

(b) exactly once?

4.60. The number of times that a person contracts a cold

in a given year is a Poisson random variable with

parameter λ = 5. Suppose that a new wonder drug

(based on large quantities of vitamin C) has just

been marketed that reduces the Poisson parame-

ter to λ = 3 for 75 percent of the population. For

the other 25 percent of the population, the drug

has no appreciable effect on colds. If an individ-

ual tries the drug for a year and has 2 colds in that

time, how likely is it that the drug is beneﬁcial for

him or her?

4.61. The probability of being dealt a full house in a

hand of poker is approximately .0014. Find an

approximation for the probability that, in 1000

hands of poker, you will be dealt at least 2 full

houses.

4.62. Consider n independent trials, each of which

results in one of the outcomes 1, ..., k with respec-

tive probabilities p

, ..., p



i=1

= 1. Show

that if all the p

are small, then the probability that

no trial outcome occurs more than once is approx-

imately equal to exp(−n(n − 1)



/2).

4.63. People enter a gambling casino at a rate of 1 every

2 minutes.

(a) What is the probability that no one enters

between 12:00 and 12:05?

(b) What is the probability that at least 4 people

enter the casino during that time?

4.64. The suicide rate in a certain state is 1 suicide per

100,000 inhabitants per month.

(a) Find the probability that, in a city of 400,000

inhabitants within this state, there will be 8 or

more suicides in a given month.

178 Chapter 4 Random Variables

(b) What is the probability that there will be at

least 2 months during the year that will have 8

or more suicides?

ber 1, what is the probability that the ﬁrst

month to have 8 or more suicides will be

month number i, i Ú 1?

What assumptions are you making?

4.65. Each of 500 soldiers in an army company indepen-

dently has a certain disease with probability 1/10

This disease will show up in a blood test, and to

facilitate matters, blood samples from all 500 sol-

diers are pooled and tested.

(a) What is the (approximate) probability that

the blood test will be positive (that is, at least

one person has the disease)?

Suppose now that the blood test yields a positive

result.

(b) What is the probability, under this circum-

stance, that more than one person has the dis-

ease?

One of the 500 people is Jones, who knows that he

has the disease.

more than one person has the disease?

Because the pooled test was positive, the author-

ities have decided to test each individual sepa-

rately. The ﬁrst i − 1 of these tests were negative,

and the ith one—which was on Jones—was posi-

tive.

(d) Given the preceding, scenario, what is the

probability, as a function of i, that any of the

remaining people have the disease?

4.66. A total of 2n people, consisting of n married cou-

ples, are randomly seated (all possible orderings

being equally likely) at a round table. Let C

denote the event that the members of couple i are

seated next to each other, i = 1, ..., n.

(a) Find P(C

(b) For j Z i, ﬁnd P(C

there are no married couples who are seated

next to each other.

4.67. Repeat the preceding problem when the seating is

random but subject to the constraint that the men

and women alternate.

4.68. In response to an attack of 10 missiles, 500 antibal-

listic missiles are launched. The missile targets of

the antiballistic missiles are independent, and each

antiballstic missile is equally likely to go towards

any of the target missiles. If each antiballistic mis-

sile independently hits its target with probability

.1, use the Poisson paradigm to approximate the

probability that all missiles are hit.

4.69. A fair coin is ﬂipped 10 times. Find the probability

that there is a string of 4 consecutive heads by

(a) using the formula derived in the text;

(b) using the recursive equations derived in the

text.

Poisson approximation.

4.70. At time 0, a coin that comes up heads with prob-

ability p is ﬂipped and falls to the ground. Sup-

pose it lands on heads. At times chosen accord-

ing to a Poisson process with rate λ, the coin is

picked up and ﬂipped. (Between these times the

coin remains on the ground.) What is the proba-

bility that the coin is on its head side at time t?

Hint What would be the conditional probability

if there were no additional ﬂips by time t, and

what would it be if there were additional ﬂips by

time t?

4.71. Consider a roulette wheel consisting of 38 num-

bers 1 through 36, 0, and double 0. If Smith always

bets that the outcome will be one of the numbers 1

through 12, what is the probability that

(a) Smith will lose his ﬁrst 5 bets;

(b) his ﬁrst win will occur on his fourth bet?

4.72. Two athletic teams play a series of games; the ﬁrst

team to win 4 games is declared the overall win-

ner. Suppose that one of the teams is stronger than

the other and wins each game with probability .6,

independently of the outcomes of the other games.

Find the probability, for i = 4, 5, 6, 7, that the

stronger team wins the series in exactly i games.

Compare the probability that the stronger team

wins with the probability that it would win a 2-out-

of-3 series.

4.73. Suppose in Problem 72 that the two teams are

evenly matched and each has probability

of win-

ning each game. Find the expected number of

games played.

4.74. An interviewer is given a list of people she can

interview. If the interviewer needs to interview 5

people, and if each person (independently) agrees

to be interviewed with probability

, what is the

probability that her list of people will enable her

to obtain her necessary number of interviews if

the list consists of (a) 5 people and (b) 8 people?

For part (b), what is the probability that the inter-

viewer will speak to exactly (c) 6 people and (d) 7

people on the list?

4.75. A fair coin is continually ﬂipped until heads

appears for the 10th time. Let X denote the num-

ber of tails that occur. Compute the probability

mass function of X.

4.76. Solve the Banach match problem (Example 8e)

when the left-hand matchbox originally contained

Theoretical Exercises 179

matches and the right-hand box contained N

matches.

4.77. In the Banach matchbox problem, ﬁnd the prob-

ability that, at the moment when the ﬁrst box is

emptied (as opposed to being found empty), the

other box contains exactly k matches.

4.78. An urn contains 4 white and 4 black balls. We ran-

domly choose 4 balls. If 2 of them are white and

2 are black, we stop. If not, we replace the balls

in the urn and again randomly select 4 balls. This

continues until exactly 2 of the 4 chosen are white.

What is the probability that we shall make exactly

n selections?

4.79. Suppose that a batch of 100 items contains 6 that

are defective and 94 that are not defective. If X

is the number of defective items in a randomly

drawn sample of 10 items from the batch, ﬁnd (a)

P{X = 0} and (b) P{X > 2}.

4.80. A game popular in Nevada gambling casinos is

Keno, which is played as follows: Twenty num-

bers are selected at random by the casino from the

set of numbers 1 through 80. A player can select

from 1 to 15 numbers; a win occurs if some frac-

tion of the player’s chosen subset matches any of

the 20 numbers drawn by the house. The payoff is a

function of the number of elements in the player’s

selection and the number of matches. For instance,

if the player selects only 1 number, then he or she

wins if this number is among the set of 20, and

the payoff is $2.2 won for every dollar bet. (As

the player’s probability of winning in this case is

, it is clear that the “fair” payoff should be $3

won for every $1 bet.) When the player selects 2

numbers, a payoff (of odds) of $12 won for every

$1 bet is made when both numbers are among

the 20,

(a) What would be the fair payoff in this case?

Let P

n, k

denote the probability that exactly

k of the n numbers chosen by the player are

among the 20 selected by the house.

(b) Compute P

n, k

selecting 10 numbers. For such a bet the

casino pays off as shown in the following

table. Compute the expected payoff:

Keno Payoffs in 10 Number Bets

Number of matches Dollars won for each $1 bet

0–4 -1

617

7 179

8 1, 299

9 2, 599

10 24, 999

4.81. In Example 8i, what percentage of i defective lots

does the purchaser reject? Find it for i = 1, 4.

Given that a lot is rejected, what is the condi-

tional probability that it contained 4 defective

components?

4.82. A purchaser of transistors buys them in lots of 20.

It is his policy to randomly inspect 4 components

from a lot and to accept the lot only if all 4 are

nondefective. If each component in a lot is, inde-

pendently, defective with probability .1, what pro-

portion of lots is rejected?

4.83. There are three highways in the county. The num-

ber of daily accidents that occur on these high-

ways are Poisson random variables with respective

parameters .3, .5, and .7. Find the expected num-

ber of accidents that will happen on any of these

highways today.

4.84. Suppose that 10 balls are put into 5 boxes, with

each ball independently being put in box i with

probability p



i=1

= 1.

(a) Find the expected number of boxes that do

not have any balls.

(b) Find the expected number of boxes that have

exactly 1 ball.

4.85. There are k types of coupons. Independently of

the types of previously collected coupons, each

new coupon collected is of type i with probability



i=1

= 1. If n coupons are collected, ﬁnd

the expected number of distinct types that appear

in this set. (That is, ﬁnd the expected number of

types of coupons that appear at least once in the

set of n coupons.)

THEORETICAL EXERCISES

4.1. There are N distinct types of coupons, and each

time one is obtained it will, independently of

past choices, be of type i with probability P

, i =

1, ..., N.LetT denote the number one need select

to obtain at least one of each type. Compute

P{T = n}.

Hint: Use an argument similar to the one used in

Example 1e.

180 Chapter 4 Random Variables

4.2. If X has distribution function F, what is the distri-

bution function of e

4.3. If X has distribution function F, what is the distri-

bution function of the random variable αX + β,

where α and β are constants, α Z 0?

4.4. For a nonnegative integer-valued random vari-

able N, show that

E[N] =



i=1

P{N Ú i}

Hint:



i=1

P{N Ú i}=



i=1



k=i

P{N = k}.Nowinter-

change the order of summation.

4.5. For a nonnegative integer-valued random variable

N, show that



i=0

iP{N > i}=

(E[N

] − E[N])

Hint:



i=0

iP{N > i}=



i=0



k=i+1

P{N = k}.Now

interchange the order of summation.

4.6. Let X be such that

P{X = 1}=p = 1 − P{X =−1}

Find c Z 1 such that E[c

] = 1.

4.7. Let X be a random variable having expected value

μ and variance σ

. Find the expected value and

variance of

Y =

X − μ

4.8. Find Var(X) if

P(X = a) = p = 1 − P(X = b)

4.9. Show how the derivation of the binomial probabil-

ities

P{X = i}=





(1 − p)

n−i

, i = 0, ..., n

leads to a proof of the binomial theorem

(x + y)



i=0





n−i

when x and y are nonnegative.

Hint: Let p =

x+y

4.10. Let X be a binomial random variable with param-

eters n and p. Show that



X + 1



1 − (1 − p)

n+1

(n + 1)p

4.11. Consider n independent sequential trials, each of

which is successful with probability p.Ifthere

is a total of k successes, show that each of the

n!/[k!(n − k)!] possible arrangements of the k suc-

cesses and n − k failures is equally likely.

4.12. There are n components lined up in a linear

arrangement. Suppose that each component inde-

pendently functions with probability p. What is the

probability that no 2 neighboring components are

both nonfunctional?

Hint: Condition on the number of defective com-

ponents and use the results of Example 4c of

Chapter 1.

4.13. Let X be a binomial random variable with param-

eters (n, p). What value of p maximizes P{X =

k}, k = 0, 1, ..., n? This is an example of a sta-

tistical method used to estimate p when a bino-

mial (n, p) random variable is observed to equal

k. If we assume that n is known, then we estimate

p by choosing that value of p which maximizes

P{X = k}. This is known as the method of maxi-

mum likelihood estimation.

4.14. A family has n

children with probability αp

, n Ú 1,

where α … (1 − p)/p.

(a) What proportion of families has no children?

(b) If each child is equally likely to be a boy or a

girl (independently of each other), what pro-

portion of families consists of k boys (and any

number of girls)?

4.15. Suppose that n independent tosses of a coin having

probability p of coming up heads are made. Show

that the probability that an even number of heads

results is

[1 + (q − p)

], where q = 1 − p.Do

this by proving and then utilizing the identity

[n/2]



i=0





n−2i

(p + q)

+ (q − p)

where [n/2] is the largest integer less than or equal

to n/2. Compare this exercise with Theoretical

Exercise 3.5 of Chapter 3.

4.16. Let X be a Poisson random variable with parame-

ter λ. Show that P{X = i} increases monotonically

and then decreases monotonically as i increases,

reaching its maximum when i is the largest integer

not exceeding λ.

Hint: Consider P{X = i}/P{X = i − 1}.

4.17. Let X be a Poisson random variable with

parameter λ.

(a) Show that

P{X is even}=

1 + e

−2λ

Theoretical Exercises 181

by using the result of Theoretical Exercise

15 and the relationship between Poisson and

binomial random variables.

(b) Verify the formula in part (a) directly by mak-

ing use of the expansion of e

−λ

+ e

4.18. Let X be a Poisson random variable with parame-

ter λ. What value of λ maximizes P{X = k}, k Ú 0?

4.19. Show that X is a Poisson random variable with

parameter λ,then

E[X

] = λE[(X + 1)

n−1

]

Now use this result to compute E[X

4.20. Consider n coins, each of which independently

comes up heads with probability p. Suppose that

n is large and p is small, and let λ = np. Suppose

that all n coins are tossed; if at least one comes

up heads, the experiment ends; if not, we again

toss all n coins, and so on. That is, we stop the

ﬁrst time that at least one of the n coins come up

heads. Let X denote the total number of heads that

appear. Which of the following reasonings con-

cerned with approximating P{X = 1} is correct

(in all cases, Y is a Poisson random variable with

parameter λ)?

(a) Because the total number of heads that occur

when all n coins are rolled is approximately a

Poisson random variable with parameter λ,

P{X = 1} L P {Y = 1}=λe

−λ

(b) Because the total number of heads that occur

when all n coins are rolled is approximately

a Poisson random variable with parameter λ,

and because we stop only when this number is

positive,

P{X = 1} L P{Y = 1|Y > 0}=

λe

−λ

1 − e

−λ

will equal 1 if none of the other n − 1 coins

come up heads. Because the number of heads

resulting from these n − 1 coins is approxi-

mately Poisson with mean (n − 1)p L λ,

P{X = 1} L P{Y = 0}=e

−λ

4.21. From a set of n randomly chosen people, let E

denote the event that persons i and j have the same

birthday. Assume that each person is equally likely

to have any of the 365 days of the year as his or her

birthday. Find

(a) P(E

3,4

1,2

);

(b) P(E

1,3

1,2

);

2,3

1,2

∩ E

1,3

What can you conclude from your answers to

parts (a)–(c) about the independence of the





events E

4.22. An urn contains 2n balls, of which 2 are numbered

1, 2 are numbered 2, ..., and 2 are numbered n.

Balls are successively withdrawn 2 at a time with-

out replacement. Let T denote the ﬁrst selection

in which the balls withdrawn have the same num-

ber (and let it equal inﬁnity if none of the pairs

withdrawn has the same number). We want to

show that, for 0 <α<1,

lim

P{T >αn}=e

−α/2

To verify the preceding formula, let M

denote the

number of pairs withdrawn in the ﬁrst k selections,

k = 1, ..., n.

(a) Argue that when n is large, M

can be

regarded as the number of successes in k

(approximately) independent trials.

(b) Approximate P{M

= 0} when n is large.

of one of the variables M

(d) Verify the limiting probability given for

P{T >αn}.

4.23. Consider a random collection of n individuals. In

approximating the probability that no 3 of these

individuals share the same birthday, a better Pois-

son approximation than that obtained in the text

(at least for values of n between 80 and 90) is

obtained by letting E

be the event that there are

at least 3 birthdays on day i, i = 1, ..., 365.

(a) Find P(E

(b) Give an approximation for the probability

that no 3 individuals share the same birthday.

can be shown to be the smallest value of n for

which the probability exceeds .5).

4.24. Here is another way to obtain a set of recur-

sive equations for determining P

, the probability

that there is a string of k consecutive heads in a

sequence of n ﬂips of a fair coin that comes up

heads with probability p:

(a) Argue that, for k < n, there will be a string of

k consecutive heads if either

1. there is a string of k consecutive heads

within the ﬁrst n − 1ﬂips,or

2. there is no string of k consecutive heads

within the ﬁrst n − k − 1ﬂips,ﬂipn − k

is a tail, and ﬂips n − k + 1, ..., n are

all heads.

(b) Using the preceding, relate P

to P

n−1

.Start-

ing with P

= p

, the recursion can be used to

obtain P

k+1

,thenP

k+1

, and so on, up to P

182 Chapter 4 Random Variables

4.25. Suppose that the number of events that occur

in a speciﬁed time is a Poisson random variable

with parameter λ. If each event is counted with

probability p, independently of every other event,

show that the number of events that are counted

is a Poisson random variable with parameter λp.

Also, give an intuitive argument as to why this

should be so. As an application of the preceding

result, suppose that the number of distinct ura-

nium deposits in a given area is a Poisson random

variable with parameter λ = 10. If, in a ﬁxed

period of time, each deposit is discovered inde-

pendently with probability

, ﬁnd the probability

that (a) exactly 1, (b) at least 1, and (c) at most 1

deposit is discovered during that time.

4.26. Prove



i=0

−λ

−x

Hint: Use integration by parts.

4.27. If X is a geometric random variable, show analyti-

cally that

P{X = n + k|X > n}=P{X = k}

Using the interpretation of a geometric random

variable, give a verbal argument as to why the pre-

ceding equation is true.

4.28. Let X be a negative binomial random variable with

parameters r and p,andletY be a binomial ran-

dom variable with parameters n and p. Show that

P{X > n}=P{Y < r}

Hint: Either one could attempt an analytical proof

of the preceding equation, which is equivalent to

proving the identity



i=n+1



i − 1

r − 1



(1 − p)

i−r

r−1



i=0





* p

(1 − p)

n−i

or one could attempt a proof that uses the prob-

abilistic interpretation of these random variables.

That is, in the latter case, start by considering a

sequence of independent trials having a common

probability p of success. Then try to express the

events {X > n } and {Y < r} in terms of the out-

comes of this sequence.

4.29. For a hypergeometric random variable, determine

P{X = k + 1}/P {X = k}

4.30. Balls numbered 1 through N are in an urn. Sup-

pose that n, n … N, of them are randomly selected

without replacement. Let Y denote the largest

number selected.

(a) Find the probability mass function of Y.

(b) Derive an expression for E[Y] and then use

Fermat’s combinatorial identity (see Theoret-

ical Exercise 11 of Chapter 1) to simplify the

expression.

4.31. A jar contains m + n chips, numbered

1, 2, ..., n + m.Asetofsizen is drawn. If we

let X denote the number of chips drawn having

numbers that exceed each of the numbers of those

remaining, compute the probability mass function

of X.

4.32. A jar contains n chips. Suppose that a boy succes-

sively draws a chip from the jar, each time replac-

ing the one drawn before drawing another. The

process continues until the boy draws a chip that

he has previously drawn. Let X denote the num-

ber of draws, and compute its probability mass

function.

4.33. Show that Equation (8.6) follows from Equation

(8.5).

4.34.

From a set of n elements, a nonempty subset is

chosen at random in the sense that all of the

nonempty subsets are equally likely to be selected.

Let X denote the number of elements in the cho-

sen subset. Using the identities given in Theoreti-

cal Exercise 12 of Chapter 1, show that

E[X] =

2 −





n−1

Var(X) =

n · 2

2n−2

− n(n + 1)2

n−2

− 1)

Show also that, for n large,

Var(X) 

in the sense that the ratio Var(X)ton/4

approaches 1 as n approaches q. Compare this

formula with the limiting form of Var(Y) when

P{Y = i}=1/n, i = 1, ..., n.

4.35. An urn initially contains one red and one blue ball.

At each stage, a ball is randomly chosen and then

replaced along with another of the same color. Let

X denote the selection number of the ﬁrst chosen

ball that is blue. For instance, if the ﬁrst selec-

tion is red and the second blue, then X is equal

to 2.

(a) Find P{X > i}, i Ú 1.

(b) Show that, with probability 1, a blue ball is

eventually chosen. (That is, show that P{X <

q}=1.)

Self-Test Problems and Exercises 183

4.36. Suppose the possible values of X are {x

}, the pos-

sible values of Y are {y

}, and the possible values

of X + Y are {z

}. Let A

denote the set of all

pairs of indices (i, j) such that x

+ y

= z

;thatis,

={(i, j) : x

+ y

= z

(a) Argue that

P{X + Y = z



(i,j)∈A

P{X = x

, Y = y

}

(b) Show that

E[X + Y] =



(i,j)∈A

+ y

)P{X = x

Y = y

}

E[X + Y] =



+ y

)P{X = x

Y = y

}

(d) Show that

P(X = x

) =



P(X = x

, Y = y

P(Y = y

) =



P{X = x

, Y = y

}

(e) Prove that

E[X + Y] = E[X] + E[Y]

SELF-TEST PROBLEMS AND EXERCISES

4.1. Suppose that the random variable X is equal to

the number of hits obtained by a certain base-

ball player in his next 3 at bats. If P{X = 1}=

.3, P{X = 2}=.2, and P{X = 0}=3P{X = 3},

ﬁnd E[X].

4.2. Suppose that X takes on one of the values 0, 1,

and 2. If for some constant c, P{X = i}=cP{X =

i − 1}, i = 1, 2, ﬁnd E[X].

4.3. A coin that, when ﬂipped, comes up heads with

probability p is ﬂipped until either heads or tails

has occurred twice. Find the expected number of

ﬂips.

4.4. A certain community is composed of m families,

of which have i children,



i=1

= m. If one of

the families is randomly chosen, let X denote the

number of children in that family. If one of the



i=1

children is randomly chosen, let Y denote

the total number of children in the family of that

child. Show that E[Y] Ú E[X].

4.5. Suppose that P{X = 0}=1 − P{X = 1}.If

E[X] = 3Var(X), ﬁnd P{X = 0}.

4.6. There are 2 coins in a bin. When one of them is

ﬂipped, it lands on heads with probability .6, and

when the other is ﬂipped, it lands on heads with

probability .3. One of these coins is to be randomly

chosen and then ﬂipped. Without knowing which

coin is chosen, you can bet any amount up to 10

dollars, and you then either win that amount if the

coincomesupheadsorloseitifitcomesuptails.

Suppose, however, that an insider is willing to sell

you, for an amount C, the information as to which

coin was selected. What is your expected payoff

if you buy this information? Note that if you buy

itandthenbetx, you will end up either winning

x − C or −x − C (that is, losing x + C in the lat-

ter case). Also, for what values of C does it pay to

purchase the information?

4.7. A philanthropist writes a positive number x on a

piece of red paper, shows the paper to an impar-

tial observer, and then turns it face down on the

table. The observer then ﬂips a fair coin. If it shows

heads, she writes the value 2x and, if tails, the value

x/2, on a piece of blue paper, which she then turns

face down on the table. Without knowing either

the value x or the result of the coin ﬂip, you have

the option of turning over either the red or the

blue piece of paper. After doing so and observing

the number written on that paper, you may elect

to receive as a reward either that amount or the

(unknown) amount written on the other piece of

paper. For instance, if you elect to turn over the

blue paper and observe the value 100, then you

can elect either to accept 100 as your reward or

to take the amount (either 200 or 50) on the red

paper. Suppose that you would like your expected

reward to be large.

(a) Argue that there is no reason to turn over the

red paper ﬁrst, because if you do so, then no

matter what value you observe, it is always

better to switch to the blue paper.

(b) Let y be a ﬁxed nonnegative value, and con-

sider the following strategy: Turn over the

blue paper, and if its value is at least y,then

accept that amount. If it is less than y,then

switch to the red paper. Let R

(x) denote the

reward obtained if the philanthropist writes

the amount x and you employ this strat-

egy. Find E[R

(x)]. Note that E[R

(x)]isthe

184 Chapter 4 Random Variables

expected reward if the philanthropist writes

the amount x when you employ the strategy

of always choosing the blue paper.

4.8. Let B(n, p) represent a binomial random variable

with parameters n and p.Arguethat

P{B(n, p) … i}=1 − P{B(n,1 − p) … n − i − 1}

Hint: The number of successes less than or equal to

i is equivalent to what statement about the number

of failures?

4.9. If X is a binomial random variable with expected

value 6 and variance 2.4, ﬁnd P{X = 5}.

4.10. An urn contains n balls numbered 1 through n.If

you withdraw m balls randomly in sequence, each

time replacing the ball selected previously, ﬁnd

P{X = k}, k = 1, ..., m,whereX is the maximum

of the m chosen numbers.

Hint: First ﬁnd P{X … k

4.11. Teams A and B play a series of games, with the

ﬁrst team to win 3 games being declared the winner

of the series. Suppose that team A independently

wins each game with probability p. Find the condi-

tional probability that team A wins

(a) the series given that it wins the ﬁrst game;

(b) the ﬁrst game given that it wins the series.

4.12. A local soccer team has 5 more games left to play.

If it wins its game this weekend, then it will play

its ﬁnal 4 games in the upper bracket of its league,

and if it loses, then it will play its ﬁnal games in

the lower bracket. If it plays in the upper bracket,

then it will independently win each of its games in

this bracket with probability .4, and if it plays in

the lower bracket, then it will independently win

each of its games with probability .7. If the proba-

bility that the team wins its game this weekend is

.5, what is the probability that it wins at least 3 of

its ﬁnal 4 games?

4.13. Each of the members of a 7-judge panel inde-

pendently makes a correct decision with probabil-

ity .7. If the panel’s decision is made by majority

rule, what is the probability that the panel makes

the correct decision? Given that 4 of the judges

agreed, what is the probability that the panel made

the correct decision?

4.14. On average, 5.2 hurricanes hit a certain region in a

year. What is the probability that there will be 3 or

fewer hurricanes hitting this year?

4.15. The number of eggs laid on a tree leaf by an insect

of a certain type is a Poisson random variable with

parameter λ. However, such a random variable

can be observed only if it is positive, since if it is

0 then we cannot know that such an insect was on

the leaf. If we let Y denote the observed number

of eggs, then

P{Y = i}=P{X = i|X > 0}

where X is Poisson with parameter λ.FindE[Y].

4.16. Each of n boys and n girls, independently and ran-

domly, chooses a member of the other sex. If a

boy and girl choose each other, they become a

couple. Number the girls, and let G

be the event

that girl number i is part of a couple. Let P

1 − P(∪

i=1

) be the probability that no couples

are formed.

(a) What is P(G

(b) What is P(G

(d) When n is large, approximate P

, the proba-

bility that exactly k couples are formed.

(e) Use the inclusion–exclusion identity to evalu-

ate P

4.17. A total of 2n people, consisting of n married cou-

ples, are randomly divided into n pairs. Arbitrarily

number the women, and let W

denote the event

that woman i is paired with her husband.

(a) Find P(W

(b) For i Z j, ﬁnd P(W

that no wife is paired with her husband.

(d) If each pairing must consist of a man and a

woman, what does the problem reduce to?

4.18. A casino patron will continue to make $5 bets on

red in roulette until she has won 4 of these bets.

(a) What is the probability that she places a total

of 9 bets?

(b) What is her expected winnings when she

stops?

Remark: On each bet, she will either win $5 with

probability

or lose $5 with probability

4.19. When three friends go for coffee, they decide who

will pay the check by each ﬂipping a coin and then

letting the “odd person” pay. If all three ﬂips pro-

duce the same result (so that there is no odd per-

son), then they make a second round of ﬂips, and

they continue to do so until there is an odd person.

What is the probability that

(a) exactly 3 rounds of ﬂips are made?

(b) more than 4 rounds are needed?

4.20. Show that if X is a geometric random variable with

parameter p,then

E[1/X] =

−p log(p)

1 − p

Hint: You will need to evaluate an expression of

the form



i=1

/i.Todoso,writea

/i =

i−1

dx,

and then interchange the sum and the integral.

4.21. Suppose that

P{X = a}=p, P{X = b}=1 − p

Self-Test Problems and Exercises 185

(a) Show that

X−b

a−b

is a Bernoulli random vari-

able.

(b) Find Var(X).

4.22. Each game you play is a win with probability p.

You plan to play 5 games, but if you win the ﬁfth

game, then you will keep on playing until you

lose.

(a) Find the expected number of games that

you play.

(b) Find the expected number of games that

you lose.

4.23. Balls are randomly withdrawn, one at a time with-

out replacement, from an urn that initially has

N white and M black balls. Find the probability

that n white balls are drawn before m black balls,

n … N, m … M.

4.24. Ten balls are to be distributed among 5 urns,

with each ball going into urn i with probabil-

ity p



i=1

= 1. Let X

denote the number of

balls that go into urn i. Assume that events cor-

responding to the locations of different balls are

independent.

(a) What type of random variable is X

?Beas

speciﬁc as possible.

(b) For i Z j, what type of random variable is

+ X

= 7}.

4.25. For the match problem (Example 5m in

Chapter 2), ﬁnd

(a) the expected number of matches.

(b) the variance of the number of matches.

4.26. Let α be the probability that a geometric random

variable X with parameter p is an even number.

(a) Find α by using the identity α =



i=1

P{X=2i}.

(b) Find α by conditioning on whether X = 1or

X > 1.

CHAPTER 5

Continuous Random Variables

5.1 INTRODUCTION

5.2 EXPECTATION AND VARIANCE OF CONTINUOUS RANDOM VARIABLES

5.3 THE UNIFORM RANDOM VARIABLE

5.4 NORMAL RANDOM VARIABLES

5.5 EXPONENTIAL RANDOM VARIABLES

5.6 OTHER CONTINUOUS DISTRIBUTIONS

5.7 THE DISTRIBUTION OF A FUNCTION OF A RANDOM VARIABLE

5.1 INTRODUCTION

In Chapter 4, we considered discrete random variables—that is, random variables

whose set of possible values is either ﬁnite or countably inﬁnite. However, there also

exist random variables whose set of possible values is uncountable. Two examples are

the time that a train arrives at a speciﬁed stop and the lifetime of a transistor. Let X

be such a random variable. We say that X is a continuous

†

random variable if there

exists a nonnegative function f , deﬁned for all real x ∈ (−q, q), having the property

that, for any set B of real numbers,

‡

P{X ∈ B}=

f (x) dx (1.1)

The function f is called the probability density function of the random variable X.

(See Figure 5.1.)

In words, Equation (1.1) states that the probability that X will be in B may be

obtained by integrating the probability density function over the set B. Since X must

assume some value, f must satisfy

1 = P{X ∈ (−q, q)}=

−q

f (x) dx

All probability statements about X can be answered in terms of f . For instance, from

Equation (1.1), letting B = [a, b], we obtain

P{a … X … b}=

f (x) dx (1.2)

†

Sometimes called absolutely continuous.

‡

Actually, for technical reasons Equation (1.1) is true only for the measurable sets B,which,

fortunately, include all sets of practical interest.

186

Section 5.1 Introduction 187

P(a ⱕ X ⱕ b) = area of shaded region

FIGURE 5.1: Probability density function f.

If we let a = b in Equation (1.2), we get

P{X = a}=

f (x) dx = 0

In words, this equation states that the probability that a continuous random variable

will assume any ﬁxed value is zero. Hence, for a continuous random variable,

P{X < a}=P{X … a}=F(a) =

−q

f (x) dx

EXAMPLE 1a

Suppose that X is a continuous random variable whose probability density function

is given by

f (x) =

C(4x − 2x

) 0 < x < 2

0 otherwise

(a) What is the value of C?

(b) Find P{X > 1}.

Solution. (a) Since f is a probability density function, we must have

−q

f (x) dx = 1,

implying that

(4x − 2x

) dx = 1



−



x=2

x=0

= 1

C =

Hence,

(b) P{X > 1}=

f (x) dx =

(4x − 2x

) dx =

188 Chapter 5 Continuous Random Variables

EXAMPLE 1b

The amount of time in hours that a computer functions before breaking down is a

continuous random variable with probability density function given by

f (x) =

λe

−x/100

x Ú 0

0 x < 0

What is the probability that

(a) a computer will function between 50 and 150 hours before breaking down?

(b) it will function for fewer than 100 hours?

Solution. (a) Since

1 =

−q

f (x) dx = λ

−x/100

we obtain

1 =−λ(100)e

−x/100

= 100λ or λ =

100

Hence, the probability that a computer will function between 50 and 150 hours before

breaking down is given by

P{50 < X < 150}=

150

100

−x/100

dx =−e

−x/100

150

= e

−1/2

− e

−3/2

L .384

(b) Similarly,

P{X < 100}=

100

−x/100

dx =−e

−x/100

100

= 1 − e

−1

L .633

In other words, approximately 63.3 percent of the time, a computer will fail before

registering 100 hours of use. .

EXAMPLE 1c

The lifetime in hours of a certain kind of radio tube is a random variable having a

probability density function given by

f (x) =

⎧

⎪

⎨

⎪

⎩

0 x … 100

100

x > 100

What is the probability that exactly 2 of 5 such tubes in a radio set will have to be

replaced within the ﬁrst 150 hours of operation? Assume that the events E

, i =

1, 2, 3, 4, 5, that the ith such tube will have to be replaced within this time are

independent.

Section 5.1 Introduction 189

Solution. From the statement of the problem, we have

P(E

) =

150

f (x) dx

= 100

150

100

−2

Hence, from the independence of the events E

, it follows that the desired probabil-

ity is











243

The relationship between the cumulative distribution F and the probability density

f is expressed by

F(a) = P{X ∈ (−q, a]}=

−q

f (x) dx

Differentiating both sides of the preceding equation yields

F(a) = f (a)

That is, the density is the derivative of the cumulative distribution function. A some-

what more intuitive interpretation of the density function may be obtained from

Equation (1.2) as follows:

a −

… X … a +

a+ε/2

a−ε/2

f (x) dx L εf (a)

when ε is small and when f (·) is continuous at x = a. In other words, the probability

that X will be contained in an interval of length ε around the point a is approximately

εf (a). From this result we see that f(a) is a measure of how likely it is that the random

variable will be near a.

EXAMPLE 1d

If X is continuous with distribution function F

and density function f

, ﬁnd the

density function of Y = 2X.

Solution. We will determine f

in two ways. The ﬁrst way is to derive, and then

differentiate, the distribution function of Y:

(a) = P{Y … a}

= P{2X … a}

= P{X … a/2}

= F

(a/2)

Differentiation gives

(a) =

(a/2)

190 Chapter 5 Continuous Random Variables

Another way to determine f

is to note that

f

(a) L P {a −



… Y … a +



}

= P{a −



… 2X … a +



}

= P{

−



… X …



}



(a/2)

Dividing through by  gives the same result as before. .

5.2 EXPECTATION AND VARIANCE OF CONTINUOUS RANDOM VARIABLES

In Chapter 4, we deﬁned the expected value of a discrete random variable X by

E[X] =



xP{X = x}

If X is a continuous random variable having probability density function f (x), then,

because

f (x) dx L P{x … X … x + dx} for dx small

it is easy to see that the analogous deﬁnition is to deﬁne the expected value of X by

E[X] =

−q

xf (x) dx

EXAMPLE 2a

Find E[X] when the density function of X is

f (x) =

2x if 0 … x … 1

0 otherwise

Solution.

E[X] =

xf (x) dx

EXAMPLE 2b

The density function of X is given by

f (x) =

1if 0… x … 1

0 otherwise

Find E[e

Section 5.2 Expectation and Variance of Continuous Random Variables 191

Solution. Let Y = e

. We start by determining F

, the probability distribution func-

tion of Y. Now, for 1 … x … e,

(x) = P{Y … x}

= P{e

… x}

= P{X … log(x)}

log(x)

f (y) dy

= log(x)

By differentiating F

(x), we can conclude that the probability density function of Y

is given by

(x) =

1 … x … e

Hence,

E[e

] = E[Y] =

−q

(x) dx

= e − 1

Although the method employed in Example 2b to compute the expected value of

a function of X is always applicable, there is, as in the discrete case, an alternative

way of proceeding. The following is a direct analog of Proposition 4.1. of Chapter 4.

Proposition 2.1. If X is a continuous random variable with probability density func-

tion f (x), then, for any real-valued function g,

E[g(X)] =

−q

g(x)f (x) dx

An application of Proposition 2.1 to Example 2b yields

E[e

] =

dx sincef (x) = 1, 0 < x < 1

= e − 1

which is in accord with the result obtained in that example.

The proof of Proposition 2.1 is more involved than that of its discrete random

variable analog. We will present such a proof under the provision that the random

variable g(X) is nonnegative. (The general proof, which follows the argument in the

case we present, is indicated in Theoretical Exercises 2 and 3.) We will need the fol-

lowing lemma, which is of independent interest.

Lemma 2.1

For a nonnegative random variable Y,

E[Y] =

P{Y > y}dy

192 Chapter 5 Continuous Random Variables

Proof. We present a proof when Y is a continuous random variable with probabil-

ity density function f

. We have

P{Y > y}dy =

(x) dx dy

where we have used the fact that P{Y > y}=

(x) dx. Interchanging the order

of integration in the preceding equation yields

P{Y > y}dy =





(x) dx

= E[Y] .

Proof of Proposition 2.1. From Lemma 2.1, for any function g for which g(x) Ú 0,

E[g(X)] =

P{g(X)>y}dy

x:g(x)>y

f (x) dx dy

x:g(x)>0

g(x)

dy f (x) dx

x:g(x)>0

g(x)f (x) dx

which completes the proof.

EXAMPLE 2c

A stick of length 1 is split at a point U that is uniformly distributed over (0, 1). Deter-

mine the expected length of the piece that contains the point p,0 … p … 1.

Solution. Let L

(U) denote the length of the substick that contains the point p,and

note that

(U) =

1 − UU< p

UU> p

(See Figure 5.2.) Hence, from Proposition 2.1,

E[L

(U)] =

(u) du

(1 − u)du +

udu

−

(1 − p)

−

+ p(1 − p)

Section 5.2 Expectation and Variance of Continuous Random Variables 193

(a)

(b)

1 – U

FIGURE 5.2: Substick containing point p:(a)U < p;(b)U > p.

Since p(1 − p) is maximized when p =

, it is interesting to note that the expected

length of the substick containing the point p is maximized when p is the midpoint of

the original stick. .

EXAMPLE 2d

Suppose that if you are s minutes early for an appointment, then you incur the cost cs,

and if you are s minutes late, then you incur the cost ks. Suppose also that the travel

time from where you presently are to the location of your appointment is a continuous

random variable having probability density function f . Determine the time at which

you should depart if you want to minimize your expected cost.

Solution. Let X denote the travel time. If you leave t minutes before your appoint-

ment, then your cost—call it C

(X)—is given by

(X) =

c(t − X) ifX … t

k(X − t) ifX Ú t

Therefore,

E[C

(X)] =

(x)f (x) dx

c(t − x)f (x) dx +

k(x − t)f (x) dx

= ct

f (x) dx − c

xf (x) dx + k

xf (x) dx − kt

f (x) dx

The value of t that minimizes E[C

(X)] can now be obtained by calculus. Differentia-

tion yields

E[C

(X)] = ct f (t) + cF(t) − ct f (t) − kt f (t) + kt f (t) − k[1 − F(t)]

= (k + c)F(t) − k

Equating the rightmost side to zero shows that the minimal expected cost is obtained

when you leave t

∗

minutes before your appointment, where t

∗

satisﬁes

F(t

∗

) =

k + c

As in Chapter 4, we can use Proposition 2.1 to show the following.

Corollary 2.1. If a and b are constants, then

E[aX + b] = aE[X] + b

194 Chapter 5 Continuous Random Variables

The proof of Corollary 2.1 for a continuous random variable X is the same as the

one given for a discrete random variable. The only modiﬁcation is that the sum is

replaced by an integral and the probability mass function by a probability density

function.

The variance of a continuous random variable is deﬁned exactly as it is for a dis-

crete random variable, namely, if X is a random variable with expected value μ,then

the variance of X is deﬁned (for any type of random variable) by

Var(X) = E[(X − μ)

]

The alternative formula,

Var(X) = E[X

] − (E[X])

is established in a manner similar to its counterpart in the discrete case.

EXAMPLE 2e

Find Var(X) for X as given in Example 2a.

Solution. We ﬁrst compute E[X

E[X

] =

−q

f (x) dx

Hence, since E[X] =

, we obtain

Var(X) =

−





It can be shown that, for constants a and b,

Var(aX + b) = a

Var(X)

The proof mimics the one given for discrete random variables.

There are several important classes of continuous random variables that appear

frequently in applications of probability; the next few sections are devoted to a study

of some of them.

5.3 THE UNIFORM RANDOM VARIABLE

A random variable is said to be uniformly distributed over the interval (0, 1) if its

probability density function is given by

f (x) =

10< x < 1

0 otherwise

(3.1)

Note that Equation (3.1) is a density function, since f (x) Ú 0and

−q

f (x) dx =

dx = 1. Because f (x)>0 only when x ∈ (0, 1), it follows that X must assume

a value in interval (0, 1). Also, since f (x) is constant for x ∈ (0, 1), X is just as likely to

Section 5.3 The Uniform Random Variable 195

␣

(

)

␤

——–

–

(a)

␣

(

)

␤

(b)

␤␣

FIGURE 5.3: Graph of (a) f(a) and (b) F(a) for a uniform (α, β) random variable.

be near any value in (0, 1) as it is to be near any other value. To verify this statement,

note that, for any 0 < a < b < 1,

P{a … X … b}=

f (x) dx = b − a

In other words, the probability that X is in any particular subinterval of (0, 1) equals

the length of that subinterval.

In general, we say that X is a uniform random variable on the interval (α, β) if the

probability density function of X is given by

f (x) =

⎧

⎪

⎨

⎪

⎩

β − α

if α<x <β

0 otherwise

(3.2)

Since F(a) =

−q

f (x) dx, it follows from Equation (3.2) that the distribution function

of a uniform random variable on the interval (α, β) is given by

F(a) =

⎧

⎪

⎨

⎪

⎩

0 a … α

a − α

β − α

α<a <β

1 a Ú β

Figure 5.3 presents a graph of f (a) and F(a).

EXAMPLE 3a

Let X be uniformly distributed over (α, β). Find (a) E[X] and (b) Var(X).

Solution. (a)

E[X] =

−q

xf (x) dx

β − α

− α

2(β − α)

β + α

196 Chapter 5 Continuous Random Variables

In words, the expected value of a random variable that is uniformly distributed

over some interval is equal to the midpoint of that interval.

(b) To ﬁnd Var(X), we ﬁrst calculate E[X

E[X

] =

β − α

− α

3(β − α)

+ αβ + α

Hence,

Var(X) =

+ αβ + α

−

(α + β)

(β − α)

Therefore, the variance of a random variable that is uniformly distributed over

some interval is the square of the length of that interval divided by 12.

EXAMPLE 3b

If X is uniformly distributed over (0, 10), calculate the probability that (a) X < 3, (b)

X > 6, and (c) 3 < X < 8.

Solution. (a) P{X < 3}=

dx =

(b) P{X > 6}=

dx =

EXAMPLE 3c

Buses arrive at a speciﬁed stop at 15-minute intervals starting at 7

A.M. That is, they

arrive at 7, 7:15, 7:30, 7:45, and so on. If a passenger arrives at the stop at a time that

is uniformly distributed between 7 and 7:30, ﬁnd the probability that he waits

(a) less than 5 minutes for a bus;

(b) more than 10 minutes for a bus.

Solution. Let X denote the number of minutes past 7 that the passenger arrives at

the stop. Since X is a uniform random variable over the interval (0, 30), it follows that

the passenger will have to wait less than 5 minutes if (and only if) he arrives between

7:10 and 7:15 or between 7:25 and 7:30. Hence, the desired probability for part (a) is

P{10 < X < 15}+P{25 < X < 30}=

dx +

dx =

Similarly, he would have to wait more than 10 minutes if he arrives between 7 and

7:05 or between 7:15 and 7:20, so the probability for part (b) is

P{0 < X < 5}+P{15 < X < 20}=

Section 5.3 The Uniform Random Variable 197

The next example was ﬁrst considered by the French mathematician Joseph

L. F. Bertrand in 1889 and is often referred to as Bertrand’s paradox. It represents

our initial introduction to a subject commonly referred to as geometrical probability.

EXAMPLE 3d

Consider a random chord of a circle. What is the probability that the length of the

chord will be greater than the side of the equilateral triangle inscribed in that circle?

Solution. As stated, the problem is incapable of solution because it is not clear what

is meant by a random chord. To give meaning to this phrase, we shall reformulate the

problem in two distinct ways.

The ﬁrst formulation is as follows: The position of the chord can be determined

by its distance from the center of the circle. This distance can vary between 0 and

r, the radius of the circle. Now, the length of the chord will be greater than the side

of the equilateral triangle inscribed in the circle if the distance from the chord to the

center of the circle is less than r/2. Hence, by assuming that a random chord is a chord

whose distance D from the center of the circle is uniformly distributed between 0 and

r, we see that the probability that the length of the chord is greater than the side of

an inscribed equilateral triangle is

D <

r/2

For our second formulation of the problem, consider an arbitrary chord of the cir-

cle; through one end of the chord, draw a tangent. The angle θ between the chord and

the tangent, which can vary from 0

◦

to 180

◦

, determines the position of the chord. (See

Figure 5.4.) Furthermore, the length of the chord will be greater than the side of the

inscribed equilateral triangle if the angle θ is between 60

◦

and 120

◦

. Hence, assuming

that a random chord is a chord whose angle θ is uniformly distributed between 0

◦

and

180

◦

, we see that the desired answer in this formulation is

P{60 <θ <120}=

120 − 60

180

Note that random experiments could be performed in such a way that

would

be the correct probability. For instance, if a circular disk of radius r is thrown on a

table ruled with parallel lines a distance 2r apart, then one and only one of these lines

would cross the disk and form a chord. All distances from this chord to the center of

the disk would be equally likely, so that the desired probability that the chord’s length

will be greater than the side of an inscribed equilateral triangle is

. In contrast, if the

␪

FIGURE 5.4

198 Chapter 5 Continuous Random Variables

experiment consisted of rotating a needle freely about a point A on the edge (see

Figure 5.4) of the circle, the desired answer would be

. .

5.4 NORMAL RANDOM VARIABLES

We say that X is a normal random variable, or simply that X is normally distributed,

with parameters μ and σ

if the density of X is given by

f (x) =

√

2πσ

−(x−μ)

/2σ

− q < x < q

This density function is a bell-shaped curve that is symmetric about μ. (See Figure 5.5.)

(b)

– 2

(a)

–3 –2 –1 0 1 2

.399

——–

␴

␮␴␮

+ 2

␴␮

–

␴␮

FIGURE 5.5: Normal density function: (a) μ = 0, σ = 1; (b) arbitrary μ, σ

The normal distribution was introduced by the French mathematician Abraham

DeMoivre in 1733, who used it to approximate probabilities associated with bino-

mial random variables when the binomial parameter n is large. This result was later

extended by Laplace and others and is now encompassed in a probability theorem

known as the central limit theorem, which is discussed in Chapter 8. The central limit

theorem, one of the two most important results in probability theory,

†

gives a theo-

retical base to the often noted empirical observation that, in practice, many random

phenomena obey, at least approximately, a normal probability distribution. Some

examples of random phenomena obeying this behavior are the height of a man, the

velocity in any direction of a molecule in gas, and the error made in measuring a

physical quantity.

To prove that f (x) is indeed a probability density function, we need to show that

√

2πσ

−q

−(x−μ)

/2σ

dx = 1

†

The other is the strong law of large numbers.

Section 5.4 Normal Random Variables 199

Making the substitution y = (x − μ)/σ , we see that

√

2πσ

−q

−(x−μ)

/2σ

dx =

√

2π

−q

−y

Hence, we must show that

−q

−y

dy =

√

2π

Toward this end, let I =

−q

−y

dy. Then

−q

−y

−q

−x

−q

−(y

)/2

dy dx

We now evaluate the double integral by means of a change of variables to polar

coordinates. (That is, let x = r cos θ, y = r sin θ ,anddy dx = rdθ dr.) Thus,

2π

−r

rdθ dr

= 2π

−r

=−2πe

−r

= 2π

Hence, I =

√

2π, and the result is proved.

An important fact about normal random variables is that if X is normally dis-

tributed with parameters μ and σ

,thenY = aX + b is normally distributed with

parameters aμ + b and a

. To prove this statement, suppose that a > 0. (The proof

when a < 0 is similar.) Let F

denote the cumulative distribution function of Y. Then

(x) = P{Y … x}

= P{aX + b … x}

= P{X …

x − b

}

= F

(

x − b

)

where F

is the cumulative distribution function of X. By differentiation, the density

function of Y is then

(x) =

(

x − b

)

√

2πaσ

exp{−(

x − b

− μ)

/2σ

}

√

2πaσ

exp{−(x − b − aμ)

/2(aσ)

}

which shows that Y is normal with parameters aμ + b and a

200 Chapter 5 Continuous Random Variables

An important implication of the preceding result is that if X is normally distributed

with parameters μ and σ

,thenZ = (X − μ)/σ is normally distributed with parame-

ters 0 and 1. Such a random variable is said to be a standard,oraunit, normal random

variable.

We now show that the parameters μ and σ

of a normal random variable represent,

respectively, its expected value and variance.

EXAMPLE 4a

Find E[X] and Var(X) when X is a normal random variable with parameters μ and σ

Solution. Let us start by ﬁnding the mean and variance of the standard normal ran-

dom variable Z = (X − μ)/σ . We have

E[Z] =

−q

(x) dx

√

2π

−q

−x

=−

√

2π

−x

−q

= 0

Thus,

Var(Z) = E[Z

]

√

2π

−q

−x

Integration by parts (with u = x and dv = xe

−x

) now gives

Var(Z) =

√

2π

(−xe

−x

−q

−x

dx)

√

2π

−q

−x

= 1

Because X = μ + σ Z, the preceding yields the results

E[X] = μ + σ E[Z] = μ

and

Var(X) = σ

Var(Z) = σ

It is customary to denote the cumulative distribution function of a standard normal

random variable by (x). That is,

(x) =

√

2π

−q

−y

The values of (x) for nonnegative x are given in Table 5.1. For negative values of x,

(x) can be obtained from the relationship

(−x) = 1 − (x) − q < x < q (4.1)

Section 5.4 Normal Random Variables 201

TABLE 5.1: AREA (x) UNDER THE STANDARD NORMAL CURVE TO THE LEFT OF X

X .00 .01 .02 .03 .04 .05 .06 .07 .08 .09

.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359

.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753

.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141

.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517

.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879

.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224

.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549

.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852

.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133

.9 .8159 .8186 .8212 .8238 .8264 .8289 .8315 .8340 .8365 .8389

1.0 .8413 .8438 .8461 .8485 .8508 .8531 .8554 .8577 .8599 .8621

1.1 .8643 .8665 .8686 .8708 .8729 .8749 .8770 .8790 .8810 .8830

1.2 .8849 .8869 .8888 .8907 .8925 .8944 .8962 .8980 .8997 .9015

1.3 .9032 .9049 .9066 .9082 .9099 .9115 .9131 .9147 .9162 .9177

1.4 .9192 .9207 .9222 .9236 .9251 .9265 .9279 .9292 .9306 .9319

1.5 .9332 .9345 .9357 .9370 .9382 .9394 .9406 .9418 .9429 .9441

1.6 .9452 .9463 .9474 .9484 .9495 .9505 .9515 .9525 .9535 .9545

1.7 .9554 .9564 .9573 .9582 .9591 .9599 .9608 .9616 .9625 .9633

1.8 .9641 .9649 .9656 .9664 .9671 .9678 .9686 .9693 .9699 .9706

1.9 .9713 .9719 .9726 .9732 .9738 .9744 .9750 .9756 .9761 .9767

2.0 .9772 .9778 .9783 .9788 .9793 .9798 .9803 .9808 .9812 .9817

2.1 .9821 .9826 .9830 .9834 .9838 .9842 .9846 .9850 .9854 .9857

2.2 .9861 .9864 .9868 .9871 .9875 .9878 .9881 .9884 .9887 .9890

2.3 .9893 .9896 .9898 .9901 .9904 .9906 .9909 .9911 .9913 .9916

2.4 .9918 .9920 .9922 .9925 .9927 .9929 .9931 .9932 .9934 .9936

2.5 .9938 .9940 .9941 .9943 .9945 .9946 .9948 .9949 .9951 .9952

2.6 .9953 .9955 .9956 .9957 .9959 .9960 .9961 .9962 .9963 .9964

2.7 .9965 .9966 .9967 .9968 .9969 .9970 .9971 .9972 .9973 .9974

2.8 .9974 .9975 .9976 .9977 .9977 .9978 .9979 .9979 .9980 .9981

2.9 .9981 .9982 .9982 .9983 .9984 .9984 .9985 .9985 .9986 .9986

3.0 .9987 .9987 .9987 .9988 .9988 .9989 .9989 .9989 .9990 .9990

3.1 .9990 .9991 .9991 .9991 .9992 .9992 .9992 .9992 .9993 .9993

3.2 .9993 .9993 .9994 .9994 .9994 .9994 .9994 .9995 .9995 .9995

3.3 .9995 .9995 .9995 .9996 .9996 .9996 .9996 .9996 .9996 .9997

3.4 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9998

The proof of Equation (4.1), which follows from the symmetry of the standard nor-

mal density, is left as an exercise. This equation states that if Z is a standard normal

random variable, then

P{Z … −x}=P{Z > x}−q < x < q

Since Z = (X − μ)/σ is a standard normal random variable whenever X is normally

distributed with parameters μ and σ

, it follows that the distribution function of X

can be expressed as

(a) = P{X … a}=P



X − μ

…

a − μ



= 



a − μ



202 Chapter 5 Continuous Random Variables

EXAMPLE 4b

If X is a normal random variable with parameters μ = 3andσ

= 9, ﬁnd (a) P{2 <

X < 5};(b)P{X > 0}; (c) P {|X − 3| > 6}.

Solution. (a)

P{2 < X < 5}=P

2 − 3

X − 3

5 − 3

= P

−

< Z <

= 





− 



−



= 





−



1 − 







L .3779

(b)

P{X > 0}=P

X − 3

0 − 3

= P{Z > −1}

= 1 − (−1)

= (1)

L .8413

(c)

P{|X − 3| > 6}=P{X > 9}+P{X < −3}

= P

X − 3

9 − 3

+ P

X − 3

−3 − 3

= P{Z > 2}+P{Z < −2}

= 1 − (2) + (−2)

= 2[1 − (2)]

L .0456

EXAMPLE 4c

An examination is frequently regarded as being good (in the sense of determining

a valid grade spread for those taking it) if the test scores of those taking the exami-

nation can be approximated by a normal density function. (In other words, a graph

of the frequency of grade scores should have approximately the bell-shaped form of

the normal density.) The instructor often uses the test scores to estimate the normal

parameters μ and σ

and then assigns the letter grade A to those whose test score

is greater than μ + σ , B to those whose score is between μ and μ + σ , C to those

whose score is between μ − σ and μ, D to those whose score is between μ − 2σ

and μ − σ , and F to those getting a score below μ − 2σ . (This strategy is sometimes

referred to as grading “on the curve.”) Since

Section 5.4 Normal Random Variables 203

P{X >μ+ σ }=P

X − μ

> 1

= 1 − (1) L .1587

P{μ<X <μ+ σ }=P

0 <

X − μ

< 1

= (1) − (0) L .3413

P{μ − σ<X <μ}=P

−1 <

X − μ

< 0

= (0) − (−1) L .3413

P{μ − 2σ<X <μ− σ }=P

−2 <

X − μ

< −1

= (2) − (1) L .1359

P{X <μ− 2σ }=P

X − μ

< −2

= (−2) L .0228

it follows that approximately 16 percent of the class will receive an A grade on the

examination, 34 percent a B grade, 34 percent a C grade, and 14 percent a D grade; 2

percent will fail. .

EXAMPLE 4d

An expert witness in a paternity suit testiﬁes that the length (in days) of human ges-

tation is approximately normally distributed with parameters μ = 270 and σ

= 100.

The defendant in the suit is able to prove that he was out of the country during a

period that began 290 days before the birth of the child and ended 240 days before

the birth. If the defendant was, in fact, the father of the child, what is the probability

that the mother could have had the very long or very short gestation indicated by the

testimony?

Solution. Let X denote the length of the gestation, and assume that the defendant

is the father. Then the probability that the birth could occur within the indicated

period is

P{X > 290orX < 240}=P{X > 290}+P {X < 240}

= P

X − 270

> 2

+ P

X − 270

< −3

= 1 − (2) + 1 − (3)

L .0241

EXAMPLE 4e

Suppose that a binary message—either 0 or 1—must be transmitted by wire from

location A to location B. However, the data sent over the wire are subject to a channel

noise disturbance, so, to reduce the possibility of error, the value 2 is sent over the

wire when the message is 1 and the value −2 is sent when the message is 0. If x, x = ;2,

is the value sent at location A,thenR, the value received at location B, is given by

R = x + N, where N is the channel noise disturbance. When the message is received

at location B, the receiver decodes it according to the following rule:

If R Ú .5, then 1 is concluded.

If R <.5, then 0 is concluded.

204 Chapter 5 Continuous Random Variables

Because the channel noise is often normally distributed, we will determine the error

probabilities when N is a standard normal random variable.

Two types of errors can occur: One is that the message 1 can be incorrectly deter-

mined to be 0, and the other is that 0 can be incorrectly determined to be 1. The ﬁrst

type of error will occur if the message is 1 and 2 + N <.5, whereas the second will

occur if the message is 0 and −2 + N Ú .5. Hence,

P{error|message is 1}=P{N < −1.5}

= 1 − (1.5) L .0668

and

P{error|message is 0}=P{N Ú 2.5}

= 1 − (2.5) L .0062

5.4.1 The Normal Approximation to the Binomial Distribution

An important result in probability theory known as the DeMoivre–Laplace limit the-

orem states that when n is large, a binomial random variable with parameters n and p

will have approximately the same distribution as a normal random variable with the

same mean and variance as the binomial. This result was proved originally for the spe-

cial case of p =

by DeMoivre in 1733 and was then extended to general p by Laplace

in 1812. It formally states that if we “standardize” the binomial by ﬁrst subtracting its

mean np and then dividing the result by its standard deviation

np(1 − p), then the

distribution function of this standardized random variable (which has mean 0 and

variance 1) will converge to the standard normal distribution function as n

→q.

The DeMoivre–Laplace limit theorem

If S

denotes the number of successes that occur when n independent trials, each

resulting in a success with probability p, are performed, then, for any a < b,

a …

− np

np(1 − p)

… b

→(b) − (a)

as n

→q.

Because the preceding theorem is only a special case of the central limit theorem,

which is presented in Chapter 8, we shall not present a proof.

Note that we now have two possible approximations to binomial probabilities: the

Poisson approximation, which is good when n is large and p is small, and the normal

approximation, which can be shown to be quite good when np(1 − p) is large. (See

Figure 5.6.) [The normal approximation will, in general, be quite good for values of n

satisfying np(1 − p) Ú 10.]

EXAMPLE 4f

Let X be the number of times that a fair coin that is ﬂipped 40 times lands on heads.

Find the probability that X = 20. Use the normal approximation and then compare

it with the exact solution.

Section 5.4 Normal Random Variables 205

0246

(10, 0.7)

810 0

0.0

0.02

0.04

0.06

0.08

0.10

0.12

0.14

0.16

0.0

0.14

0.12

0.10

0.08

0.06

0.04

0.02

0.0

0.30

0.25

0.20

0.15

0.10

0.05

0.0

0.20

0.15

0.10

0.05

10 15

(30, 0.7)

20 25 30 0 10 20

(50, 0.7)

30 40 50

510

(20, 0.7)

15 20

FIGURE 5.6: The probability mass function of a binomial (n, p) random variable becomes more and more

“normal” as n becomes larger and larger.

Solution. To employ the normal approximation, note that because the binomial is

a discrete integer-valued random variable, whereas the normal is a continuous ran-

dom variable, it is best to write P{X = i} as P{i − 1/2 < X < i + 1/2} before

applying the normal approximation (this is called the continuity correction). Doing

so gives

P{X = 20}=P{19.5 … X < 20.5}

= P

19.5 − 20

√

X − 20

√

20.5 − 20

√

L P

−.16 <

X − 20

√

<.16

L (.16) − (−.16) L .1272

The exact result is

P{X = 20}=







L .1254

EXAMPLE 4g

The ideal size of a ﬁrst-year class at a particular college is 150 students. The college,

knowing from past experience that, on the average, only 30 percent of those accepted

for admission will actually attend, uses a policy of approving the applications of 450

students. Compute the probability that more than 150 ﬁrst-year students attend this

college.

Solution. If X denotes the number of students that attend, then X is a binomial ran-

dom variable with parameters n = 450 and p = .3. Using the continuity correction,

206 Chapter 5 Continuous Random Variables

we see that the normal approximation yields

P{X Ú 150.5}=P

X − (450)(.3)

√

450(.3)(.7)

150.5 − (450)(.3)

√

450(.3)(.7)

L 1 − (1.59)

L .0559

Hence, less than 6 percent of the time do more than 150 of the ﬁrst 450 accepted

actually attend. (What independence assumptions have we made?) .

EXAMPLE 4h

To determine the effectiveness of a certain diet in reducing the amount of cholesterol

in the bloodstream, 100 people are put on the diet. After they have been on the diet

for a sufﬁcient length of time, their cholesterol count will be taken. The nutritionist

running this experiment has decided to endorse the diet if at least 65 percent of the

people have a lower cholesterol count after going on the diet. What is the probability

that the nutritionist endorses the new diet if, in fact, it has no effect on the cholesterol

level?

Solution. Let us assume that if the diet has no effect on the cholesterol count, then,

strictly by chance, each person’s count will be lower than it was before the diet with

probability

. Hence, if X is the number of people whose count is lowered, then the

probability that the nutritionist will endorse the diet when it actually has no effect on

the cholesterol count is

100



i=65



100





100

= P{X Ú 64.5}

= P

⎧

⎪

⎨

⎪

⎩

X − (100)(

)

100(

)(

)

Ú 2.9

⎫

⎪

⎬

⎪

⎭

L 1 − (2.9)

L .0019

EXAMPLE 4i

Fifty-two percent of the residents of New York City are in favor of outlawing cigarette

smoking in publicly owned areas. Approximate the probability that more than 50

percent of a random sample of n people from New York are in favor of this prohibi-

tion when

(a) n = 11

(b) n = 101

How large would n have to be to make this probability exceed .95?

Solution. Let N denote the number of residents of New York City. To answer the

preceding question, we must ﬁrst understand that a random sample of size n is a

sample such that the n people were chosen in such a manner that each of the





subsets of n people had the same chance of being the chosen subset. Consequently,

Section 5.4 Normal Random Variables 207

, the number of people in the sample who are in favor of the smoking prohibition,

is a hypergeometric random variable. That is, S

has the same distribution as the

number of white balls obtained when n balls are chosen from an urn of N balls, of

which .52N are white. But because N and .52N are both large in comparison with the

sample size n, it follows from the binomial approximation to the hypergeometric (see

Section 4.8.3) that the distribution of S

is closely approximated by a binomial dis-

tribution with parameters n and p = .52. The normal approximation to the binomial

distribution then shows that

P{S

>.5n}=P

− .52n

√

n(.52)(.48)

.5n − .52n

√

n(.52)(.48)

= P

− .52n

√

n(.52)(.48)

> −.04

√

L (.04

√

Thus,

P{S

>.5n} L

⎧

⎨

⎩

(.1328) = .5528, if n = 11

(.4020) = .6562, if n = 101

(1.2665) = .8973, if n = 1001

In order for this probability to be at least .95, we would need (.04

√

n)>.95. Because

(x) is an increasing function and (1.645) = .95, this means that

.04

√

n > 1.645

n Ú 1691.266

That is, the sample size would have to be at least 1692. .

Historical Notes Concerning the Normal Distribution

The normal distribution was introduced by the French mathematician Abraham

DeMoivre in 1733. DeMoivre, who used this distribution to approximate proba-

bilities connected with coin tossing, called it the exponential bell-shaped curve. Its

usefulness, however, became truly apparent only in 1809, when the famous German

mathematician Karl Friedrich Gauss used it as an integral part of his approach to

predicting the location of astronomical entities. As a result, it became common after

this time to call it the Gaussian distribution.

During the mid- to late 19th century, however, most statisticians started to

believe that the majority of data sets would have histograms conforming to the

Gaussian bell-shaped form. Indeed, it came to be accepted that it was “normal”

for any well-behaved data set to follow this curve. As a result, following the lead of

the British statistician Karl Pearson, people began referring to the Gaussian curve

by calling it simply the normal curve. (A partial explanation as to why so many data

sets conform to the normal curve is provided by the central limit theorem, which is

presented in Chapter 8.)

Abraham DeMoivre (1667–1754)

Today there is no shortage of statistical consultants, many of whom ply their trade

in the most elegant of settings. However, the ﬁrst of their breed worked, in the early

208 Chapter 5 Continuous Random Variables

years of the 18th century, out of a dark, grubby betting shop in Long Acres, London,

known as Slaughter’s Coffee House. He was Abraham DeMoivre, a Protestant

refugee from Catholic France, and, for a price, he would compute the probability of

gambling bets in all types of games of chance.

Although DeMoivre, the discoverer of the normal curve, made his living at the

coffee shop, he was a mathematician of recognized abilities. Indeed, he was a mem-

ber of the Royal Society and was reported to be an intimate of Isaac Newton.

Listen to Karl Pearson imagining DeMoivre at work at Slaughter’s Coffee

House: “I picture DeMoivre working at a dirty table in the coffee house with a broken-

down gambler beside him and Isaac Newton walking through the crowd to his corner

to fetch out his friend. It would make a great picture for an inspired artist.”

Karl Friedrich Gauss

Karl Friedrich Gauss (1777–1855), one of the earliest users of the normal curve,

was one of the greatest mathematicians of all time. Listen to the words of the

well-known mathematical historian E. T. Bell, as expressed in his 1954 book Men

of Mathematics: In a chapter entitled “The Prince of Mathematicians,” he writes,

“Archimedes, Newton, and Gauss; these three are in a class by themselves among

the great mathematicians, and it is not for ordinary mortals to attempt to rank them

in order of merit. All three started tidal waves in both pure and applied mathemat-

ics. Archimedes esteemed his pure mathematics more highly than its applications;

Newton appears to have found the chief justiﬁcation for his mathematical inventions

in the scientiﬁc uses to which he put them; while Gauss declared it was all one to him

whether he worked on the pure or on the applied side.”

‡

5.5 EXPONENTIAL RANDOM VARIABLES

A continuous random variable whose probability density function is given, for some

λ>0, by

f (x) =

λe

−λx

if x Ú 0

0ifx < 0

is said to be an exponential random variable (or, more simply, is said to be exponen-

tially distributed) with parameter λ. The cumulative distribution function F(a) of an

exponential random variable is given by

F(a) = P{X … a}

λe

−λx

=−e

−λx

= 1 − e

−λa

a Ú 0

Note that F(q) =

λe

−λx

dx = 1, as, of course, it must. The parameter λ will now

be shown to equal the reciprocal of the expected value.

EXAMPLE 5a

Let X be an exponential random variable with parameter λ. Calculate (a) E[X]and

(b) Var(X).

Section 5.5 Exponential Random Variables 209

Solution. (a) Since the density function is given by

f (x) =

λe

−λx

x Ú 0

0 x < 0

we obtain, for n > 0,

E[X

] =

λe

−λx

Integrating by parts (with λe

−λx

= dv and u = x

) yields

E[X

] =−x

−λx

n−1

= 0 +

λe

−λx

n−1

E[X

n−1

]

Letting n = 1andthenn = 2 gives

E[X] =

E[X

] =

E[X] =

(b) Hence,

Var(X) =

−





Thus, the mean of the exponential is the reciprocal of its parameter λ, and the vari-

ance is the mean squared. .

In practice, the exponential distribution often arises as the distribution of the

amount of time until some speciﬁc event occurs. For instance, the amount of time

(starting from now) until an earthquake occurs, or until a new war breaks out, or

until a telephone call you receive turns out to be a wrong number are all random

variables that tend in practice to have exponential distributions. (For a theoretical

explanation of this phenomenon, see Section 4.7.)

EXAMPLE 5b

Suppose that the length of a phone call in minutes is an exponential random variable

with parameter λ =

. If someone arrives immediately ahead of you at a public

telephone booth, ﬁnd the probability that you will have to wait

(a) more than 10 minutes;

(b) between 10 and 20 minutes.

Solution. Let X denote the length of the call made by the person in the booth. Then

the desired probabilities are

(a)

P{X > 10}=1 − F(10)

= e

−1

L .368

210 Chapter 5 Continuous Random Variables

(b)

P{10 < X < 20}=F(20) − F(10)

= e

−1

− e

−2

L .233

We say that a nonnegative random variable X is memoryless if

P{X > s + t | X > t}=P{X > s} for all s, t Ú 0 (5.1)

If we think of X as being the lifetime of some instrument, Equation (5.1) states that

the probability that the instrument survives for at least s +t hours, given that it has

survived t hours, is the same as the initial probability that it survives for at least s

hours. In other words, if the instrument is alive at age t, the distribution of the remain-

ing amount of time that it survives is the same as the original lifetime distribution.

(That is, it is as if the instrument does not “remember” that it has already been in use

foratimet.)

Equation (5.1) is equivalent to

P{X > s + t, X > t}

P{X > t}

= P{X > s}

P{X > s + t}=P{X > s}P{X > t} (5.2)

Since Equation (5.2) is satisﬁed when X is exponentially distributed (for e

−λ(s+t)

−λs

−λt

), it follows that exponentially distributed random variables are memoryless.

EXAMPLE 5c

Consider a post ofﬁce that is staffed by two clerks. Suppose that when Mr. Smith

enters the system, he discovers that Ms. Jones is being served by one of the clerks

and Mr. Brown by the other. Suppose also that Mr. Smith is told that his service will

begin as soon as either Ms. Jones or Mr. Brown leaves. If the amount of time that

a clerk spends with a customer is exponentially distributed with parameter λ, what

is the probability that, of the three customers, Mr. Smith is the last to leave the post

ofﬁce?

Solution. The answer is obtained by reasoning as follows: Consider the time at which

Mr. Smith ﬁrst ﬁnds a free clerk. At this point, either Ms. Jones or Mr. Brown would

have just left, and the other one would still be in service. However, because the expo-

nential is memoryless, it follows that the additional amount of time that this other

person (either Ms. Jones or Mr. Brown) would still have to spend in the post ofﬁce is

exponentially distributed with parameter λ. That is, it is the same as if service for that

person were just starting at this point. Hence, by symmetry, the probability that the

remaining person ﬁnishes before Smith leaves must equal

. .

It turns out that not only is the exponential distribution memoryless, but it is also

the unique distribution possessing this property. To see this, suppose that X is mem-

oryless and let

F(x) = P{X > x}. Then, by Equation (5.2),

F(s + t) = F(s)F(t)

That is,

F(·) satisﬁes the functional equation

g(s + t) = g(s)g(t)

Section 5.5 Exponential Random Variables 211

However, it turns out that the only right continuous solution of this functional

equation is

†

g(x) = e

−λx

(5.3)

and, since a distribution function is always right continuous, we must have

F(x) = e

−λx

or F(x) = P{X … x}=1 − e

−λx

which shows that X is exponentially distributed.

EXAMPLE 5d

Suppose that the number of miles that a car can run before its battery wears out is

exponentially distributed with an average value of 10,000 miles. If a person desires to

take a 5000-mile trip, what is the probability that he or she will be able to complete the

trip without having to replace the car battery? What can be said when the distribution

is not exponential?

Solution. It follows by the memoryless property of the exponential distribution that

the remaining lifetime (in thousands of miles) of the battery is exponential with

parameter λ =

. Hence, the desired probability is

P{remaining lifetime > 5}=1 − F(5) = e

−5λ

= e

−1/2

L .604

However, if the lifetime distribution F is not exponential, then the relevant probabil-

ity is

P{lifetime > t + 5|lifetime > t}=

1 − F(t + 5)

1 − F(t)

where t is the number of miles that the battery had been in use prior to the start of

the trip. Therefore, if the distribution is not exponential, additional information is

needed (namely, the value of t) before the desired probability can be calculated. .

A variation of the exponential distribution is the distribution of a random variable

that is equally likely to be either positive or negative and whose absolute value is

exponentially distributed with parameter λ,

λ Ú 0. Such a random variable is said to

have a Laplace distribution,

‡

and its density is given by

f (x) =

λe

−λ|x|

− q < x < q

†

One can prove Equation (5.3) as follows: If g(s + t) = g(s)g(t),then





= g





= g





and repeating this yields g(m/n) = g

(1/n).Also,

g(1) = g



+ ··· +



= g





or g





= (g(1))

1/n

Hence, g(m/n) = (g(1))

m/n

,which,sinceg is right continuous, implies that g (x) = (g(1))

Because g(1) =









Ú 0, we obtain g(x) = e

−λx

,whereλ =−log(g(1)).

‡

It also is sometimes called the double exponential random variable.

212 Chapter 5 Continuous Random Variables

Its distribution function is given by

F(x) =

⎧

⎪

⎨

⎪

⎩

−q

λe

λx

dx x < 0

−q

λe

λx

dx +

λe

−λx

dx x > 0

⎧

⎨

⎩

λx

x < 0

1 −

−λx

x > 0

EXAMPLE 5e

Consider again Example 4e, which supposes that a binary message is to be transmit-

ted from A to B, with the value 2 being sent when the message is 1 and −2 when it is

0. However, suppose now that, rather than being a standard normal random variable,

the channel noise N is a Laplacian random variable with parameter λ = 1. Suppose

again that if R is the value received at location B, then the message is decoded as

follows:

If R Ú .5, then 1 is concluded.

If R <.5, then 0 is concluded.

In this case, where the noise is Laplacian with parameter λ = 1, the two types of

errors will have probabilities given by

P{error|message 1 is sent}=P{N < −1.5}

−1.5

L .1116

P{error|message 0 is sent}=P{N Ú 2.5}

−2.5

L .041

On comparing this with the results of Example 4e, we see that the error probabilities

are higher when the noise is Laplacian with λ = 1 than when it is a standard normal

variable.

5.5.1 Hazard Rate Functions

Consider a positive continuous random variable X that we interpret as being the

lifetime of some item. Let X have distribution function F and density f . The hazard

rate (sometimes called the failure rate) function λ(t) of F is deﬁned by

λ(t) =

f (t)

F(t)

, where

F = 1 − F

To interpret λ(t), suppose that the item has survived for a time t and we desire the

probability that it will not survive for an additional time dt. That is, consider P{X ∈

(t, t + dt)|X > t}. Now,

Section 5.5 Exponential Random Variables 213

P{X ∈ (t, t + dt)|X > t}=

P{X ∈ (t, t + dt), X > t}

P{X > t}

P{X ∈ (t, t + dt)}

P{X > t}

f (t)

F(t)

Thus, λ(t) represents the conditional probability intensity that a t-unit-old item will fail.

Suppose now that the lifetime distribution is exponential. Then, by the memoryless

property, it follows that the distribution of remaining life for a t-year-old item is the

same as that for a new item. Hence, λ(t) should be constant. In fact, this checks out,

since

λ(t) =

f (t)

F(t)

λe

−λt

= λ

Thus, the failure rate function for the exponential distribution is constant. The param-

eter λ is often referred to as the rate of the distribution.

It turns out that the failure rate function λ(t) uniquely determines the distribu-

tion F. To prove this, note that, by deﬁnition,

λ(t) =

F(t)

1 − F(t)

Integrating both sides yields

log(1 − F(t)) =−

λ(t) dt + k

1 − F(t) = e

exp

−

λ(t) dt

Letting t = 0 shows that k = 0; thus,

F(t) = 1 − exp

−

λ(t) dt

(5.4)

Hence, a distribution function of a positive continuous random variable can be

speciﬁed by giving its hazard rate function. For instance, if a random variable has a

linear hazard rate function—that is, if

λ(t) = a + bt

then its distribution function is given by

F(t) = 1 − e

−at−bt

and differentiation yields its density, namely,

214 Chapter 5 Continuous Random Variables

f (t) = (a + bt)e

−(at+bt

/2)

t Ú 0

When a = 0, the preceding equation is known as the Rayleigh density function.

EXAMPLE 5f

One often hears that the death rate of a person who smokes is, at each age, twice that

of a nonsmoker. What does this mean? Does it mean that a nonsmoker has twice the

probability of surviving a given number of years as does a smoker of the same age?

Solution. If λ

(t) denotes the hazard rate of a smoker of age t and λ

(t) that of a

nonsmoker of age t, then the statement at issue is equivalent to the statement that

(t) = 2λ

(t)

The probability that an A-year-old nonsmoker will survive until age B, A < B,is

P{A-year-old nonsmoker reaches age B}

= P{nonsmoker’s lifetime > B|nonsmoker’s lifetime > A}

1 − F

non

(B)

1 − F

non

(A)

exp

−

(t) dt

exp

−

(t) dt

from (5.4)

= exp

−

(t) dt

whereas the corresponding probability for a smoker is, by the same reasoning,

P{A-year-old smoker reaches age B}=exp

−

(t) dt

= exp

−2

(t) dt

⎡

⎣

exp

−

(t) dt

⎤

⎦

In other words, for two people of the same age, one of whom is a smoker and

the other a nonsmoker, the probability that the smoker survives to any given age

is the square (not one-half) of the corresponding probability for a nonsmoker. For

instance, if λ

(t) =

,50 … t … 60, then the probability that a 50-year-old nonsmoker

reaches age 60 is e

−1/3

L .7165, whereas the corresponding probability for a smoker

is e

−2/3

L .5134. .

Section 5.6 Other Continuous Distributions 215

5.6 OTHER CONTINUOUS DISTRIBUTIONS

5.6.1 The Gamma Distribution

A random variable is said to have a gamma distribution with parameters (α, λ), λ>0,

α>0, if its density function is given by

f (x) =

⎧

⎪

⎨

⎪

⎩

λe

−λx

(λx)

α−1

(α)

x Ú 0

0 x < 0

where (α), called the gamma function, is deﬁned as

(α) =

−y

α−1

Integration of (α) by parts yields

(α) =−e

−y

α−1

−y

(α − 1)y

α−2

= (α − 1)

−y

α−2

dy (6.1)

= (α − 1)(α − 1)

For integral values of α, say, α = n, we obtain, by applying Equation (6.1) repeatedly,

(n) = (n − 1)(n − 1)

= (n − 1)(n − 2)(n − 2)

=···

= (n − 1)(n − 2) ···3 · 2(1)

Since (1) =

−x

dx = 1, it follows that, for integral values of n,

(n) = (n − 1)!

When α is a positive integer, say, α = n, the gamma distribution with parameters

(α, λ) often arises, in practice as the distribution of the amount of time one has to

wait until a total of n events has occurred. More speciﬁcally, if events are occurring

randomly and in accordance with the three axioms of Section 4.7, then it turns out

that the amount of time one has to wait until a total of n events has occurred will be a

gamma random variable with parameters (n, λ). To prove this, let T

denote the time

at which the nth event occurs, and note that T

is less than or equal to t if and only

if the number of events that have occurred by time t is at least n. That is, with N(t)

equal to the number of events in [0, t],

P{T

… t}=P{N(t) Ú n}



j=n

P{N(t) = j}



j=n

−λt

(λt)

216 Chapter 5 Continuous Random Variables

where the ﬁnal identity follows because the number of events in [0, t]hasa

Poisson distribution with parameter λt. Differentiation of the preceding now yields

the density function of T

f (t) =



j=n

−λt

j(λt )

j−1

−



j=n

λe

−λt

(λt)



j=n

λe

−λt

(λt)

j−1

(j − 1)!

−



j=n

λe

−λt

(λt)

λe

−λt

(λt)

n−1

(n − 1)!

Hence, T

has the gamma distribution with parameters (n, λ). (This distribution is

often referred to in the literature as the n-Erlang distribution.) Note that when n = 1,

this distribution reduces to the exponential distribution.

The gamma distribution with λ =

and α = n/2, n a positive integer, is called the

(read “chi-squared”) distribution with n degrees of freedom. The chi-squared dis-

tribution often arises in practice as the distribution of the error involved in attempt-

ing to hit a target in n-dimensional space when each coordinate error is normally

distributed. This distribution will be studied in Chapter 6, where its relation to the

normal distribution is detailed.

EXAMPLE 6a

Let X be a gamma random variable with parameters α and λ. Calculate (a) E[X]and

(b) Var(X).

Solution. (a)

E[X] =

(α)

λxe

−λx

(λx)

α−1

λ(α)

λe

−λx

(λx)

(α + 1)

λ(α)

by Equation (6.1)

(b) By ﬁrst calculating E[X

], we can show that

Var(X) =

The details are left as an exercise. .

5.6.2 The Weibull Distribution

The Weibull distribution is widely used in engineering practice due to its versatility. It

was originally proposed for the interpretation of fatigue data, but now its use has been

extended to many other engineering problems. In particular, it is widely used in the

ﬁeld of life phenomena as the distribution of the lifetime of some object, especially

when the “weakest link” model is appropriate for the object. That is, consider an

Section 5.6 Other Continuous Distributions 217

object consisting of many parts, and suppose that the object experiences death (fail-

ure) when any of its parts fail. It has been shown (both theoretically and empirically)

that under these conditions a Weibull distribution provides a close approximation to

the distribution of the lifetime of the item.

The Weibull distribution function has the form

F(x) =

⎧

⎪

⎨

⎪

⎩

0 x … ν

1 − exp

−



x − ν



x >ν

(6.2)

A random variable whose cumulative distribution function is given by Equation (6.2)

is said to be a Weibull random variable with parameters ν, α,andβ. Differentiation

yields the density:

f (x) =

⎧

⎪

⎨

⎪

⎩

0 x … ν



x − ν



β−1

exp

−



x − ν



x >ν

5.6.3 The Cauchy Distribution

A random variable is said to have a Cauchy distribution with parameter θ, −q <θ<

q, if its density is given by

f (x) =

1 + (x − θ)

− q < x < q

EXAMPLE 6b

Suppose that a narrow-beam ﬂashlight is spun around its center, which is located a

unit distance from the x-axis. (See Figure 5.7.) Consider the point X at which the

beam intersects the x-axis when the ﬂashlight has stopped spinning. (If the beam is

not pointing toward the x-axis, repeat the experiment.)

␪

0 Xx-axis

FIGURE 5.7

As indicated in Figure 5.7, the point X is determined by the angle θ between the

ﬂashlight and the y-axis, which, from the physical situation, appears to be uniformly

distributed between −π/2andπ/2. The distribution function of X is thus given by

F(x) = P{X … x}

= P{tan θ … x}

= P{θ … tan

−1

tan

−1

218 Chapter 5 Continuous Random Variables

where the last equality follows since θ, being uniform over (−π/2, π/2),has

distribution

P{θ … a}=

a − (−π/2)

−

< a <

Hence, the density function of X is given by

f (x) =

F(x) =

π(1 + x

)

− q < x < q

and we see that X has the Cauchy distribution.

†

5.6.4 The Beta Distribution

A random variable is said to have a beta distribution if its density is given by

f (x) =

⎧

⎪

⎨

⎪

⎩

B(a, b)

a−1

(1 − x)

b−1

0 < x < 1

0 otherwise

where

B(a, b) =



a−1

(1 − x)

b−1

The beta distribution can be used to model a random phenomenon whose set of

possible values is some ﬁnite interval [c, d]—which, by letting c denote the origin and

taking d − c as a unit measurement, can be transformed into the interval [0, 1].

When a = b, the beta density is symmetric about

, giving more and more weight

to regions about

as the common value a increases. (See Figure 5.8.) When b > a,

the density is skewed to the left (in the sense that smaller values become more likely);

and it is skewed to the right when a > b. (See Figure 5.9.)

The relationship

B(a, b) =

(a)(b)

(a + b)

(6.3)

can be shown to exist between

B(a, b) =



a−1

(1 − x)

b−1

and the gamma function.

†

That

(tan

−1

x) = 1/(1 + x

) can be seen as follows: If y = tan

−1

x,thentany = x,so

1 =

(tan y) =

(tan y)



sin y

cos y





cos

y + sin

cos



cos

sin

y + cos

tan

y + 1

+ 1

Section 5.7 The Distribution of a Function of a Random Variable 219

(x)

–

a =

a = 3

a = 10

a = 1

FIGURE 5.8: Beta densities with parameters (a, b)whena = b.

(x)

–

a =

a = 6

FIGURE 5.9: Beta densities with parameters (a, b)whena/(a + b) = 1/20.

Upon using Equation (6.1) along with the identity (6.3), it is an easy matter to show

that if X is a beta random variable with parameters a and b,then

E[X] =

a + b

Var(X) =

(a + b)

(a + b + 1)

Remark. A veriﬁcation of Equation (6.3) appears in Example 7c of Chapter 6. .

5.7 THE DISTRIBUTION OF A FUNCTION OF A RANDOM VARIABLE

Often, we know the probability distribution of a random variable and are interested

in determining the distribution of some function of it. For instance, suppose that we

know the distribution of X and want to ﬁnd the distribution of g(X).Todoso,itis

220 Chapter 5 Continuous Random Variables

necessary to express the event that g(X) … y in terms of X being in some set. We

illustrate with the following examples.

EXAMPLE 7a

Let X be uniformly distributed over (0, 1). We obtain the distribution of the random

variable Y, deﬁned by Y = X

, as follows: For 0 … y … 1,

(y) = P{Y … y}

= P{X

… y}

= P{X … y

1/n

}

= F

1/n

)

= y

1/n

For instance, the density function of Y is given by

(y) =

⎧

⎪

⎨

⎪

⎩

1/n−1

0 … y … 1

0 otherwise

EXAMPLE 7b

If X is a continuous random variable with probability density f

, then the distribution

of Y = X

is obtained as follows: For y Ú 0,

(y) = P{Y … y}

= P{X

… y}

= P{−

√

y … X …

√

= F

(

√

y) − F

(−

√

Differentiation yields

(y) =

√

(

√

y) + f

(−

√

y)] .

EXAMPLE 7c

If X has a probability density f

,thenY =|X| has a density function that is obtained

as follows: For y Ú 0,

(y) = P{Y … y}

= P{|X| … y}

= P{−y … X … y}

= F

(y) − F

(−y)

Hence, on differentiation, we obtain

(y) = f

(y) + f

(−y) y Ú 0

Section 5.7 The Distribution of a Function of a Random Variable 221

The method employed in Examples 7a through 7c can be used to prove

Theorem 7.1.

Theorem 7.1. Let X be a continuous random variable having probability density func-

tion f

. Suppose that g(x) is a strictly monotonic (increasing or decreasing), differen-

tiable (and thus continuous) function of x. Then the random variable Y deﬁned by

Y = g(X) has a probability density function given by

(y) =

⎧

⎪

⎨

⎪

⎩

−1

(y)]

−1

(y)

if y = g(x) for some x

0 if y Z g(x) for all x

where g

−1

(y) is deﬁned to equal that value of x such that g(x) = y.

We shall prove Theorem 7.1 when g(x) is an increasing function.

Proof. Suppose that y = g(x) for some x. Then, with Y = g(X),

(y) = P{g(X) … y}

= P{X … g

−1

(y)}

= F

−1

(y))

Differentiation gives

(y) = f

−1

(y))

−1

(y) .

which agrees with Theorem 7.1, since g

−1

(y) is nondecreasing, so its derivative is non-

negative.

When y Z g(x) for any x,thenF

(y) is either 0 or 1, and in either case f

(y) = 0.

EXAMPLE 7d

Let X be a continuous nonnegative random variable with density function f , and let

Y = X

. Find f

, the probability density function of Y.

Solution. If g(x) = x

,then

−1

(y) = y

1/n

and

−1

(y)}=

1/n−1

Hence, from Theorem 7.1, we obtain

(y) =

1/n−1

f (y

1/n

)

For n = 2, this gives

(y) =

√

f (

√

which (since X Ú 0) is in agreement with the result of Example 7b. .

222 Chapter 5 Continuous Random Variables

SUMMARY

A random variable X is continuous if there is a nonnegative function f , called the

probability density function of X, such that, for any set B,

P{X ∈ B}=

f (x) dx

If X is continuous, then its distribution function F will be differentiable and

F(x) = f (x)

The expected value of a continuous random variable X is deﬁned by

E[X] =

−q

xf (x) dx

A useful identity is that, for any function g,

E[g(X)] =

−q

g(x)f (x) dx

As in the case of a discrete random variable, the variance of X is deﬁned by

Var(X) = E[(X − E[X])

]

A random variable X is said to be uniform over the interval (a, b) if its probability

density function is given by

f (x) =

⎧

⎪

⎨

⎪

⎩

b − a

a … x … b

0 otherwise

Its expected value and variance are

E[X] =

a + b

Var(X) =

(b − a)

A random variable X is said to be normal with parameters μ and σ

if its probability

density function is given by

f (x) =

√

2πσ

−(x−μ)

/2σ

− q < x < q

It can be shown that

μ = E[X] σ

= Var(X)

If X is normal with mean μ and variance σ

,thenZ, deﬁned by

Z =

X − μ

is normal with mean 0 and variance 1. Such a random variable is said to be a standard

normal random variable. Probabilities about X can be expressed in terms of proba-

bilities about the standard normal variable Z, whose probability distribution function

can be obtained either from Table 5.1 or from a website.

Summary 223

When n is large, the probability distribution function of a binomial random vari-

able with parameters n and p can be approximated by that of a normal random

variable having mean np and variance np(1 − p).

A random variable whose probability density function is of the form

f (x) =

λe

−λx

x Ú 0

0 otherwise

is said to be an exponential random variable with parameter λ. Its expected value and

variance are, respectively,

E[X] =

Var(X) =

A key property possessed only by exponential random variables is that they are mem-

oryless, in the sense that, for positive s and t,

P{X > s + t|X > t}=P{X > s}

If X represents the life of an item, then the memoryless property states that, for any

t, the remaining life of a t-year-old item has the same probability distribution as the

life of a new item. Thus, one need not remember the age of an item to know its

distribution of remaining life.

Let X be a nonnegative continuous random variable with distribution function F

and density function f . The function

λ(t) =

f (t)

1 − F(t)

t Ú 0

is called the hazard rate,orfailure rate, function of F. If we interpret X as being the

life of an item, then, for small values of dt, λ(t) dt is approximately the probability

that a t-unit-old item will fail within an additional time dt.IfF is the exponential

distribution with parameter λ,then

λ(t) = λ t Ú 0

In addition, the exponential is the unique distribution having a constant failure rate.

A random variable is said to have a gamma distribution with parameters α and λ

if its probability density function is equal to

f (x) =

λe

−λx

(λx)

α−1

(α)

x Ú 0

and is 0 otherwise. The quantity (α) is called the gamma function and is deﬁned by

(α) =

−x

α−1

The expected value and variance of a gamma random variable are, respectively,

E[X] =

Var(X) =

224 Chapter 5 Continuous Random Variables

A random variable is said to have a beta distribution with parameters (a, b)ifits

probability density function is equal to

f (x) =

B(a, b)

a−1

(1 − x)

b−1

0 … x … 1

and is equal to 0 otherwise. The constant B(a, b) is given by

B(a, b) =

a−1

(1 − x)

b−1

The mean and variance of such a random variable are, respectively,

E[X] =

a + b

Var(X) =

(a + b)

(a + b + 1)

PROBLEMS

5.1. Let X be a random variable with probability den-

sity function

f (x) =

c(1 − x

) −1 < x < 1

0 otherwise

(a) What is the value of c?

(b) What is the cumulative distribution function

of X?

5.2. A system consisting of one original unit plus a

spare can function for a random amount of time X.

If the density of X is given (in units of months) by

f (x) =

Cxe

−x/2

x > 0

0 x … 0

what is the probability that the system functions

for at least 5 months?

5.3. Consider the function

f (x) =

C(2x − x

) 0 < x <

0 otherwise

Could f be a probability density function? If so,

determine C. Repeat if f (x) were given by

f (x) =

C(2x − x

) 0 < x <

0 otherwise

5.4. The probability density function of X, the lifetime

of a certain type of electronic device (measured in

hours), is given by

f (x) =

⎧

⎪

⎨

⎪

⎩

x > 10

0 x … 10

(a) Find P{X > 20}.

(b) What is the cumulative distribution function

of X?

devices, at least 3 will function for at least 15

hours? What assumptions are you making?

5.5. A ﬁlling station is supplied with gasoline once a

week. If its weekly volume of sales in thousands of

gallons is a random variable with probability den-

sity function

f (x) =

5(1 − x)

0 < x < 1

0 otherwise

what must the capacity of the tank be so that the

probability of the supply’s being exhausted in a

given week is .01?

5.6. Compute E[X]ifX has a density function given by

(a) f (x) =

⎧

⎪

⎨

⎪

⎩

−x/2

x > 0

0 otherwise

;

(b) f (x) =

c(1 − x

) −1 < x < 1

0 otherwise

;

⎧

⎪

⎨

⎪

⎩

x > 5

0 x … 5

Problems 225

5.7. The density function of X is given by

f (x) =

a + bx

0 … x … 1

0 otherwise

If E[X] =

, ﬁnd a and b.

5.8. The lifetime in hours of an electronic tube is a ran-

dom variable having a probability density function

given by

f (x) = xe

−x

x Ú 0

Compute the expected lifetime of such a tube.

5.9. Consider Example 4b of Chapter 4, but now sup-

pose that the seasonal demand is a continuous ran-

dom variable having probability density function f .

Show that the optimal amount to stock is the value

∗

that satisﬁes

F(s

∗

) =

b + 

where b is net proﬁt per unit sale,  is the net loss

per unit unsold, and F is the cumulative distribu-

tion function of the seasonal demand.

5.10. Trains headed for destination A arrive at the train

station at 15-minute intervals starting at 7

A.M.,

whereas trains headed for destination B arrive at

15-minute intervals starting at 7:05

A.M.

(a) If a certain passenger arrives at the station at

a time uniformly distributed between 7 and

A.M. and then gets on the ﬁrst train that

arrives, what proportion of time does he or

she go to destination A?

(b) What if the passenger arrives at a time uni-

formly distributed between 7:10 and 8:10

A.M.?

5.11. A point is chosen at random on a line segment

of length L. Interpret this statement, and ﬁnd the

probability that the ratio of the shorter to the

longer segment is less than

5.12. A bus travels between the two cities A and B,

which are 100 miles apart. If the bus has a break-

down, the distance from the breakdown to city A

has a uniform distribution over (0, 100). There is a

bus service station in city A,inB,andinthecen-

ter of the route between A and B. It is suggested

that it would be more efﬁcient to have the three

stations located 25, 50, and 75 miles, respectively,

from A. Do you agree? Why?

5.13. You arrive at a bus stop at 10 o’clock, knowing

that the bus will arrive at some time uniformly dis-

tributed between 10 and 10:30.

(a) What is the probability that you will have to

wait longer than 10 minutes?

(b) If, at 10:15, the bus has not yet arrived, what

is the probability that you will have to wait at

least an additional 10 minutes?

5.14. Let X be a uniform (0, 1) random variable. Com-

pute E[X

] by using Proposition 2.1, and then

check the result by using the deﬁnition of expec-

tation.

5.15. If X is a normal random variable with parameters

μ = 10 and σ

= 36, compute

(a) P{X > 5};

(b) P{4 < X < 16};

(d) P{X < 20};

(e) P{X > 16}.

5.16. The annual rainfall (in inches) in a certain region is

normally distributed with μ = 40 and σ = 4. What

is the probability that, starting with this year, it will

take over 10 years before a year occurs having a

rainfall of over 50 inches? What assumptions are

you making?

5.17. A man aiming at a target receives 10 points if his

shot is within 1 inch of the target, 5 points if it is

between 1 and 3 inches of the target, and 3 points if

it is between 3 and 5 inches of the target. Find the

expected number of points scored if the distance

from the shot to the target is uniformly distributed

between 0 and 10.

5.18. Suppose that X is a normal random variable with

mean 5. If P{X > 9}=.2, approximately what is

Var(X)?

5.19. Let X be a normal random variable with mean

12 and variance 4. Find the value of c such that

P{X > c}=.10.

5.20. If 65 percent of the population of a large commu-

nity is in favor of a proposed rise in school taxes,

approximate the probability that a random sample

of 100 people will contain

(a) at least 50 who are in favor of the proposition;

(b) between 60 and 70 inclusive who are in favor;

5.21. Suppose that the height, in inches, of a 25-year-old

man is a normal random variable with parameters

μ = 71 and σ

= 6.25. What percentage of 25-

year-old men are over 6 feet, 2 inches tall? What

percentage of men in the 6-footer club are over 6

feet, 5 inches?

5.22. The width of a slot of a duralumin forging is (in

inches) normally distributed with μ = .9000 and

σ = .0030. The speciﬁcation limits were given as

.9000 ; .0050.

(a) What percentage of forgings will be defec-

tive?

226 Chapter 5 Continuous Random Variables

(b) What is the maximum allowable value of σ

that will permit no more than 1 in 100 defec-

tives when the widths are normally distributed

with μ = .9000 and σ ?

5.23. One thousand independent rolls of a fair die will

be made. Compute an approximation to the prob-

ability that the number 6 will appear between 150

and 200 times inclusively. If the number 6 appears

exactly 200 times, ﬁnd the probability that the

number 5 will appear less than 150 times.

5.24. The lifetimes of interactive computer chips pro-

duced by a certain semiconductor manufacturer

are normally distributed with parameters μ =

1.4 * 10

hours and σ = 3 * 10

hours. What is the

approximate probability that a batch of 100 chips

will contain at least 20 whose lifetimes are less than

1.8 * 10

5.25. Each item produced by a certain manufacturer is,

independently, of acceptable quality with proba-

bility .95. Approximate the probability that at most

10 of the next 150 items produced are unaccept-

able.

5.26. Two types of coins are produced at a factory: a fair

coin and a biased one that comes up heads 55 per-

cent of the time. We have one of these coins, but

do not know whether it is a fair coin or a biased

one. In order to ascertain which type of coin we

have, we shall perform the following statistical test:

We shall toss the coin 1000 times. If the coin lands

on heads 525 or more times, then we shall conclude

that it is a biased coin, whereas if it lands on heads

less than 525 times, then we shall conclude that it

is a fair coin. If the coin is actually fair, what is the

probability that we shall reach a false conclusion?

What would it be if the coin were biased?

5.27. In 10,000 independent tosses of a coin, the coin

landed on heads 5800 times. Is it reasonable to

assume that the coin is not fair? Explain.

5.28. Twelve percent of the population is left handed.

Approximate the probability that there are at least

20 left-handers in a school of 200 students. State

your assumptions.

5.29. A model for the movement of a stock supposes

that if the present price of the stock is s, then, after

one period, it will be either us with probability p

or ds with probability 1 − p. Assuming that suc-

cessive movements are independent, approximate

the probability that the stock’s price will be up

at least 30 percent after the next 1000 periods if

u = 1.012, d = 0.990, and p = .52.

5.30. An image is partitioned into two regions, one

white and the other black. A reading taken from

a randomly chosen point in the white section will

give a reading that is normally distributed with μ =

4andσ

= 4, whereas one taken from a randomly

chosen point in the black region will have a nor-

mally distributed reading with parameters (6, 9).

A point is randomly chosen on the image and has

a reading of 5. If the fraction of the image that is

black is α, for what value of α would the probabil-

ity of making an error be the same, regardless of

whether one concluded that the point was in the

black region or in the white region?

5.31. (a) A ﬁre station is to be located along a road of

length A, A < q. If ﬁres occur at points uni-

formly chosen on (0, A), where should the sta-

tion be located so as to minimize the expected

distance from the ﬁre? That is, choose a so

as to

minimize E[|X − a|]

when X is uniformly distributed over (0, A).

(b) Now suppose that the road is of inﬁnite

length—stretching from point 0 outward to q.

If the distance of a ﬁre from point 0 is expo-

nentially distributed with rate λ, where should

the ﬁre station now be located? That is, we

want to minimize E[|X − a|], where X is now

exponential with rate λ.

5.32. The time (in hours) required to repair a machine is

an exponentially distributed random variable with

parameter λ =

. What is

(a) the probability that a repair time exceeds 2

hours?

(b) the conditional probability that a repair takes

at least 10 hours, given that its duration

exceeds 9 hours?

5.33. The number of years a radio functions is exponen-

tially distributed with parameter λ =

. If Jones

buys a used radio, what is the probability that it

will be working after an additional 8 years?

5.34. Jones ﬁgures that the total number of thousands

of miles that an auto can be driven before it would

need to be junked is an exponential random vari-

able with parameter

. Smith has a used car that

he claims has been driven only 10,000 miles. If

Jones purchases the car, what is the probability

that she would get at least 20,000 additional miles

out of it? Repeat under the assumption that the

lifetime mileage of the car is not exponentially dis-

tributed, but rather is (in thousands of miles) uni-

formly distributed over (0, 40).

5.35. The lung cancer hazard rate λ(t) of a t-year-old

male smoker is such that

λ(t) = .027 + .00025(t − 40)

t Ú 40

Assuming that a 40-year-old male smoker survives

all other hazards, what is the probability that he

survives to (a) age 50 and (b) age 60 without con-

tracting lung cancer?

Theoretical Exercises 227

5.36. Suppose that the life distribution of an item has the

hazard rate function λ(t) = t

, t > 0. What is the

probability that

(a) the item survives to age 2?

(b) the item’s lifetime is between .4 and 1.4?

5.37. If X is uniformly distributed over (−1, 1), ﬁnd

(a) P{|X| >

};

(b) the density function of the random variable

|X|.

5.38. If Y is uniformly distributed over (0, 5), what

is the probability that the roots of the equation

+ 4xY + Y + 2 = 0 are both real?

5.39. If X is an exponential random variable with

parameter λ = 1, compute the probability den-

sity function of the random variable Y deﬁned by

Y = log X.

5.40. If X is uniformly distributed over (0, 1), ﬁnd the

density function of Y = e

5.41. Find the distribution of R = A sin θ,whereA is

a ﬁxed constant and θ is uniformly distributed on

(−π/2, π/2). Such a random variable R arises in

the theory of ballistics. If a projectile is ﬁred from

the origin at an angle α from the earth with a speed

ν, then the point R at which it returns to the earth

can be expressed as R = (v

/g) sin 2α,whereg is

the gravitational constant, equal to 980 centime-

ters per second squared.

THEORETICAL EXERCISES

5.1. The speed of a molecule in a uniform gas at equi-

librium is a random variable whose probability

density function is given by

f (x) =

−bx

x Ú 0

0 x < 0

where b = m/2kT and k, T,andm denote, respec-

tively, Boltzmann’s constant, the absolute temper-

ature of the gas, and the mass of the molecule.

Evaluate a in terms of b.

5.2. Show that

E[Y] =

P{Y > y}dy −

P{Y < −y}dy

Hint: Show that

P{Y < −y}dy =−

−q

(x) dx

P{Y > y}dy =

(x) dx

5.3. Show that if X has density function f ,then

E[g(X)] =

−q

g(x)f (x) dx

Hint: Using Theoretical Exercise 2, start with

E[g(X)] =

P{g(X)>y}dy −

P{g(X)<−y}dy

and then proceed as in the proof given in the text

when g(X) Ú 0.

5.4. Prove Corollary 2.1.

5.5. Use the result that, for a nonnegative random vari-

able Y,

E[Y] =

P{Y > t}dt

to show that, for a nonnegative random vari-

able X,

E[X

] =

n−1

P{X > x}dx

Hint: Start with

E[X

] =

P{X

> t}dt

and make the change of variables t = x

5.6. Deﬁne a collection of events E

,0 < a < 1, hav-

ing the property that P(E

) = 1foralla but







= 0.

Hint:LetX be uniform over (0, 1) and deﬁne each

in terms of X.

5.7. The standard deviation of X, denoted SD(X),is

given by

SD(X) =

Var(X)

Find SD(aX + b) if X has variance σ

5.8. Let X be a random variable that takes on values

between 0 and c.Thatis,P{0 … X … c}=1.

Show that

Var(X) …

Hint: One approach is to ﬁrst argue that

E[X

] … cE[X]

228 Chapter 5 Continuous Random Variables

and then use this inequality to show that

Var(X) … c

[α(1 − α)]whereα =

E[X]

5.9. Show that Z is a standard normal random variable,

then, for x > 0,

(a) P{Z > x}=P{Z < −x};

(b) P{|Z| > x}=2P{Z > x};

5.10. Let f (x) denote the probability density function of

a normal random variable with mean μ and vari-

ance σ

. Show that μ − σ and μ + σ are points

of inﬂection of this function. That is, show that



(x) = 0whenx = μ − σ or x = μ + σ .

5.11. Let Z be a standard normal random variable Z,

and let g be a differentiable function with deriva-

tive g



(a) Show that E[g



(Z)] = E[Zg(Z)]

(b) Show that E[Z

n+1

] = nE[Z

n−1

]

5.12. Use the identity of Theoretical Exercise 5 to derive

E[X

]whenX is an exponential random variable

with parameter λ.

5.13. The median of a continuous random variable hav-

ing distribution function F is that value m such that

F(m) =

. That is, a random variable is just as

likely to be larger than its median as it is to be

smaller. Find the median of X if X is

(a) uniformly distributed over (a, b);

(b) normal with parameters μ, σ

;

5.14. The mode of a continuous random variable having

density f is the value of x for which f (x) attains its

maximum. Compute the mode of X in cases (a),

(b), and (c) of Theoretical Exercise 5.13.

5.15. If X is an exponential random variable with

parameter λ,andc > 0, show that cX is exponen-

tial with parameter λ/c.

5.16. Compute the hazard rate function of X when X is

uniformly distributed over (0, a).

5.17. If X has hazard rate function λ

(t), compute the

hazard rate function of aX where a is a positive

constant.

5.18. Verify that the gamma density function integrates

to 1.

5.19. If X is an exponential random variable with mean

1/λ, show that

E[X

] =

k = 1, 2, ...

Hint: Make use of the gamma density function to

evaluate the preceding.

5.20. Verify that

Var(X) =

when X is a gamma random variable with param-

eters α and λ.

5.21. Show that 





√

π.

Hint: 





−x

−1/2

dx. Make the change

of variables y =

√

2x and then relate the resulting

expression to the normal distribution.

5.22. Compute the hazard rate function of a gamma ran-

dom variable with parameters (α, λ) and show it is

increasing when α Ú 1 and decreasing when α … 1.

5.23. Compute the hazard rate function of a Weibull

random variable and show it is increasing when

β Ú 1 and decreasing when β … 1.

5.24. Show that a plot of log(log(1 − F(x))

−1

) against

log x will be a straight line with slope β when

F(·) is a Weibull distribution function. Show also

that approximately 63.2 percent of all observa-

tions from such a distribution will be less than α.

Assume that v = 0.

5.25. Let

Y =



X − ν



Show that if X is a Weibull random variable with

parameters ν, α,andβ,thenY is an exponential

random variable with parameter λ = 1andvice

versa.

5.26. If X is a beta random variable with parameters a

and b, show that

E[X] =

a + b

Var(X) =

(a + b)

(a + b + 1)

5.27. If X is uniformly distributed over (a, b), what ran-

dom variable, having a linear relation with X,is

uniformly distributed over (0, 1)?

5.28. Consider the beta distribution with parameters

(a, b). Show that

(a) when a > 1andb > 1, the density is uni-

modal (that is, it has a unique mode) with

mode equal to (a − 1)/(a + b − 2);

(b) when a … 1, b … 1, and a + b < 2, the den-

sity is either unimodal with mode at 0 or 1 or

U-shaped with modes at both 0 and 1;

5.29. Let X be a continuous random variable having

cumulative distribution function F. Deﬁne the ran-

dom variable Y by Y = F(X). Show that Y is

uniformly distributed over (0, 1).

5.30. Let X have probability density f

. Find the prob-

ability density function of the random variable Y

deﬁned by Y = aX + b.

Self-Test Problems and Exercises 229

5.31. Find the probability density function of Y = e

when X is normally distributed with parameters μ

and σ

. The random variable Y is said to have a

lognormal distribution (since log Y has a normal

distribution) with parameters μ and σ

5.32. Let X and Y be independent random vari-

ables that are both equally likely to be either

1, 2, ..., (10)

,whereN is very large. Let D denote

the greatest common divisor of X and Y,andlet

= P{D = k}.

(a) Give a heuristic argument that Q

Hint: Note that in order for D to equal k, k

must divide both X and Y and also X/k,and

Y/k must be relatively prime. (That is, X/k,

and Y/k must have a greatest common divisor

equal to 1.)

(b) Use part (a) to show that

= P{X and Y are relatively prime}



k=1

1/k

It is a well-known identity that



1/k

/6, so Q

= 6/π

. (In number theory, this

is known as the Legendre theorem.)



i=1



− 1



where P

is the ith-smallest prime greater

than 1.

Hint: X and Y will be relatively prime if they

have no common prime factors. Hence, from

part (b), we see that



i=1



− 1



which was noted without explanation in

Problem 11 of Chapter 4. (The relationship

between this problem and Problem 11 of

Chapter 4 is that X and Y are relatively prime

if XY has no multiple prime factors.)

5.33. Prove Theorem 7.1 when g(x) is a decreasing func-

tion.

SELF-TEST PROBLEMS AND EXERCISES

5.1. The number of minutes of playing time of a certain

high school basketball player in a randomly cho-

sen game is a random variable whose probability

density function is given in the following ﬁgure:

.025

.050

10 20 30 40

Find the probability that the player plays

(a) over 15 minutes;

(b) between 20 and 35 minutes;

(d) more than 36 minutes.

5.2. For some constant c, the random variable X has

the probability density function

f (x) =

0 < x < 1

0 otherwise

Find (a) c and (b) P{X > x},0 < x < 1.

5.3. For some constant c, the random variable X has

the probability density function

f (x) =

0 < x < 2

0 otherwise

Find (a) E[X]and(b)Var(X).

5.4. The random variable X has the probability density

function

f (x) =

ax + bx

0 < x < 1

0 otherwise

If E[X] = .6, ﬁnd (a) P{X <

} and (b) Var(X).

5.5. The random variable X is said to be a discrete uni-

form random variable on the integers 1, 2, ..., n if

P{X = i}=

i = 1, 2, ..., n

For any nonnegative real number x,letInt(x)

(sometimes written as [x]) be the largest inte-

ger that is less than or equal to x. Show that

U is a uniform random variable on (0, 1), fain

230 Chapter 5 Continuous Random Variables

X = Int(nU) + 1 is a discrete uniform random

variable on 1, ..., n.

5.6. Your company must make a sealed bid for a con-

struction project. If you succeed in winning the

contract (by having the lowest bid), then you plan

to pay another ﬁrm 100 thousand dollars to do

the work. If you believe that the minimum bid

(in thousands of dollars) of the other participating

companies can be modeled as the value of a ran-

dom variable that is uniformly distributed on (70,

140), how much should you bid to maximize your

expected proﬁt?

5.7. To be a winner in a certain game, you must be

successful in three successive rounds. The game

depends on the value of U, a uniform random vari-

able on (0, 1). If U >.1, then you are successful

in round 1; if U >.2, then you are successful in

round 2; and if U >.3, then you are successful in

round 3.

(a) Find the probability that you are successful in

round 1.

(b) Find the conditional probability that you are

successful in round 2 given that you were suc-

cessful in round 1.

successful in round 3 given that you were suc-

cessful in rounds 1 and 2.

(d) Find the probability that you are a winner.

5.8. A randomly chosen IQ test taker obtains a score

that is approximately a normal random variable

with mean 100 and standard deviation 15. What is

the probability that the score of such a person is

(a) above 125; (b) between 90 and 110?

5.9. Suppose that the travel time from your home to

your ofﬁce is normally distributed with mean 40

minutes and standard deviation 7 minutes. If you

want to be 95 percent certain that you will not be

late for an ofﬁce appointment at 1

P.M., what is the

latest time that you should leave home?

5.10. The life of a certain type of automobile tire is nor-

mally distributed with mean 34,000 miles and stan-

dard deviation 4000 miles.

(a) What is the probability that such a tire lasts

over 40,000 miles?

(b) What is the probability that it lasts between

30,000 and 35,000 miles?

is the conditional probability that the tire sur-

vives another 10,000 miles?

5.11. The annual rainfall in Cleveland, Ohio is approx-

imately a normal random variable with mean 40.2

inches and standard deviation 8.4 inches. What is

the probability that

(a) next year’s rainfall will exceed 44 inches?

(b) the yearly rainfalls in exactly 3 of the next 7

years will exceed 44 inches?

Assume that if A

is the event that the rainfall

exceeds 44 inches in year i (from now), then the

events A

, i Ú 1, are independent.

5.12. The following table uses 1992 data concerning the

percentages of male and female full-time workers

whose annual salaries fall into different ranges:

Earnings range Percentage Percentage

of females of males

…9999 8.6 4.4

10,000–19,999 38.0 21.1

20,000–24,999 19.4 15.8

25,000–49,999 29.2 41.5

Ú50,000 4.8 17.2

Suppose that random samples of 200 male and 200

female full-time workers are chosen. Approximate

the probability that

(a) at least 70 of the women earn $25,000 or more;

(b) at most 60 percent of the men earn $25,000

or more;

half the women earn $20,000 or more.

5.13. At a certain bank, the amount of time that a cus-

tomer spends being served by a teller is an expo-

nential random variable with mean 5 minutes. If

there is a customer in service when you enter the

bank, what is the probability that he or she will still

be with the teller after an additional 4 minutes?

5.14. Suppose that the cumulative distribution function

of the random variable X is given by

F(x) = 1 − e

−x

x > 0

Evaluate (a) P{X > 2};(b)P{1 < X < 3};(c)the

hazard rate function of F;(d)E[X]; (e) Var(X).

Hint: For parts (d) and (e), you might want to

make use of the results of Theoretical Exercise 5.

5.15. The number of years that a washing machine func-

tions is a random variable whose hazard rate func-

tion is given by

λ(t) =

⎧

⎨

⎩

.20< t < 2

.2 + .3(t − 2) 2 … t < 5

1.1 t > 5

(a) What is the probability that the machine will

still be working 6 years after being purchased?

(b) If it is still working 6 years after being pur-

chased, what is the conditional probability

that it will fail within the next 2 years?

5.16. A standard Cauchy random variable has density

function

f (x) =

π(1 + x

)

− q < x < q

Self-Test Problems and Exercises 231

Show that X is a standard Cauchy random

variable, then 1/X is also a standard Cauchy ran-

dom variable.

5.17. A roulette wheel has 38 slots, numbered 0, 00, and

1 through 36. If you bet 1 on a speciﬁed num-

ber then you either win 35 if the roulette ball

lands on that number or lose 1 if it does not.

If you continually make such bets, approximate

the probability that

(a) you are winning after 34 bets;

(b) you are winning after 1000 bets;

Assume that each roll of the roulette ball is equally

likely to land on any of the 38 numbers.

5.18. There are two types of batteries in a bin. When in

use, type i batteries last (in hours) an exponentially

distributed time with rate λ

, i = 1, 2. A battery

that is randomly chosen from the bin will be a type

i battery with probability p



i=1

= 1. If a ran-

domly chosen battery is still operating after t hours

of use, what is the probability that it will still be

operating after an additional s hours?

5.19. Evidence concerning the guilt or innocence of a

defendant in a criminal investigation can be sum-

marized by the value of an exponential random

variable X whose mean μ depends on whether the

defendant is guilty. If innocent, μ = 1; if guilty,

μ = 2. The deciding judge will rule the defen-

dant guilty if X > c for some suitably chosen

value of c.

(a) If the judge wants to be 95 percent certain

that an innocent man will not be convicted,

what should be the value of c?

(b) Using the value of c found in part (a), what is

the probability that a guilty defendant will be

convicted?

5.20. For any real number y, deﬁne y

y,ify Ú 0

0, if y < 0

Let c be a constant.

(a) Show that

E[(Z − c)

] =

√

2π

−c

− c(1 − (c))

when Z is a standard normal random variable.

(b) Find E[(X − c)

]whenX is normal with

mean μ and variance σ

CHAPTER 6

Jointly Distributed Random Variables

6.1 JOINT DISTRIBUTION FUNCTIONS

6.2 INDEPENDENT RANDOM VARIABLES

6.3 SUMS OF INDEPENDENT RANDOM VARIABLES

6.4 CONDITIONAL DISTRIBUTIONS: DISCRETE CASE

6.5 CONDITIONAL DISTRIBUTIONS: CONTINUOUS CASE

6.6 ORDER STATISTICS

6.7 JOINT PROBABILITY DISTRIBUTION OF FUNCTIONS OF RANDOM VARIABLES

6.8 EXCHANGEABLE RANDOM VARIABLES

6.1 JOINT DISTRIBUTION FUNCTIONS

Thus far, we have concerned ourselves only with probability distributions for single

random variables. However, we are often interested in probability statements con-

cerning two or more random variables. In order to deal with such probabilities, we

deﬁne, for any two random variables X and Y,thejoint cumulative probability distri-

bution function of X and Y by

F(a, b) = P{X … a, Y … b}−q < a, b < q

The distribution of X can be obtained from the joint distribution of X and Y as

follows:

(a) = P{X … a}

= P{X … a, Y < q}

= P



lim

b→q

{X … a, Y … b}



= lim

b→q

P{X … a, Y … b}

= lim

b→q

F(a, b)

K F(a, q)

Note that, in the preceding set of equalities, we have once again made use of the fact

that probability is a continuous set (that is, event) function. Similarly, the cumulative

distribution function of Y is given by

(b) = P{Y … b}

= lim

a→q

F(a, b)

K F(q, b)

232

Section 6.1 Joint Distribution Functions 233

The distribution functions F

and F

are sometimes referred to as the marginal dis-

tributions of X and Y.

All joint probability statements about X and Y can, in theory, be answered in terms

of their joint distribution function. For instance, suppose we wanted to compute the

joint probability that X is greater than a and Y is greater than b. This could be done

as follows:

P{X > a, Y > b}=1 − P({X > a, Y > b}

)

= 1 − P({X > a}

∪{Y > b}

)

= 1 − P({X … a}∪{Y … b}) (1.1)

= 1 − [P{X … a}+P{Y … b}−P{X … a, Y … b}]

= 1 − F

(a) − F

(b) + F(a, b)

Equation (1.1) is a special case of the following equation, whose veriﬁcation is left as

an exercise:

P{a

< X … a

, b

< Y … b

}

= F(a

, b

) + F(a

, b

) − F(a

, b

) − F(a

, b

) (1.2)

whenever a

< a

, b

< b

In the case when X and Y are both discrete random variables, it is convenient to

deﬁne the joint probability mass function of X and Y by

p(x, y) = P{X = x, Y = y}

The probability mass function of X can be obtained from p(x, y) by

(x) = P{X = x}



y:p(x,y)>0

p(x, y)

Similarly,

(y) =



x:p(x,y)>0

p(x, y)

EXAMPLE 1a

Suppose that 3 balls are randomly selected from an urn containing 3 red, 4 white, and

5 blue balls. If we let X and Y denote, respectively, the number of red and white balls

chosen, then the joint probability mass function of X and Y, p(i, j) = P{X = i, Y = j},

is given by

p(0, 0) =











220

p(0, 1) =













220

p(0, 2) =













220

p(0, 3) =











220

234 Chapter 6 Jointly Distributed Random Variables

p(1, 0) =













220

p(1, 1) =













220

p(1, 2) =













220

p(2, 0) =













220

p(2, 1) =













220

p(3, 0) =











220

These probabilities can most easily be expressed in tabular form, as in Table 6.1. The

reader should note that the probability mass function of X is obtained by computing

the row sums, whereas the probability mass function of Y is obtained by comput-

ing the column sums. Because the individual probability mass functions of X and Y

thus appear in the margin of such a table, they are often referred to as the marginal

probability mass functions of X and Y, respectively. .

TABLE 6.1: P{X = i, Y = j}

i 0123Rowsum= P{X = i}

220

108

220

000

220

Column sum = P{Y = j}

220

112

220

EXAMPLE 1b

Suppose that 15 percent of the families in a certain community have no children, 20

percent have 1 child, 35 percent have 2 children, and 30 percent have 3. Suppose

further that in each family each child is equally likely (independently) to be a boy or

a girl. If a family is chosen at random from this community, then B, the number of

boys, and G, the number of girls, in this family will have the joint probability mass

function shown in Table 6.2.

Section 6.1 Joint Distribution Functions 235

TABLE 6.2: P{B = i, G = j}

i 0123Rowsum= P{B = i}

0 .15 .10 .0875 .0375 .3750

1 .10 .175 .1125 0 .3875

2 .0875 .1125 0 0 .2000

3 .0375 0 0 0 .0375

Columnsum = P{G = j} .3750 .3875 .2000 .0375

The probabilities shown in Table 6.2 are obtained as follows:

P{B = 0, G = 0}=P{no children}=.15

P{B = 0, G = 1}=P{1 girl and total of 1 child}

= P{1 child}P{1 girl|1 child}=(.20)





P{B = 0, G = 2}=P{2 girls and total of 2 children}

= P{2 children}P{2 girls|2 children}=(.35)





We leave the veriﬁcation of the remaining probabilities in the table to the reader. .

We say that X and Y are jointly continuous if there exists a function f (x, y), deﬁned

for all real x and y, having the property that, for every set C of pairs of real numbers

(that is, C is a set in the two-dimensional plane),

P{(X, Y) ∈ C}=

(x,y)∈C

f (x, y) dx dy (1.3)

The function f (x, y) is called the joint probability density function of X and Y.IfA

and B are any sets of real numbers, then, by deﬁning C ={(x, y) : x ∈ A, y ∈ B},we

see from Equation (1.3) that

P{X ∈ A, Y ∈ B}=

f (x, y) dx dy (1.4)

Because

F(a, b) = P{X ∈ (−q, a], Y ∈ (−q, b]}

−q

f (x, y) dx dy

it follows, upon differentiation, that

f (a, b) =

∂

∂a ∂b

F(a, b)

wherever the partial derivatives are deﬁned. Another interpretation of the joint den-

sity function, obtained from Equation (1.4), is

236 Chapter 6 Jointly Distributed Random Variables

P{a < X < a + da, b < Y < b + db}=

d+db

a+da

f (x, y) dx dy

L f (a, b) da db

when da and db are small and f (x, y) is continuous at a, b. Hence, f (a, b) is a measure

of how likely it is that the random vector (X, Y) will be near (a, b).

If X and Y are jointly continuous, they are individually continuous, and their prob-

ability density functions can be obtained as follows:

P{X ∈ A}=P{X ∈ A, Y ∈ (−q, q)}

−q

f (x, y) dy dx

(x) dx

where

(x) =

−q

f (x, y) dy

is thus the probability density function of X. Similarly, the probability density func-

tion of Y is given by

(y) =

−q

f (x, y) dx

EXAMPLE 1c

The joint density function of X and Y is given by

f (x, y) =

−x

−2y

0 < x < q,0< y < q

0 otherwise

Compute (a) P{X > 1, Y < 1},(b)P{X < Y}, and (c) P {X < a}.

Solution. (a)

P{X > 1, Y < 1}=

−x

−2y

dx dy

−2y



−e

−x



= e

−1

−2y

= e

−1

(1 − e

−2

)

(b)

P{X < Y}=

(x,y):x<y

−x

−2y

dx dy

−x

−2y

dx dy

Section 6.1 Joint Distribution Functions 237

−2y

(1 − e

−y

)dy

−2y

dy −

−3y

= 1 −

(c)

P{X < a}=

−2y

−x

dy dx

−x

= 1 − e

−a

EXAMPLE 1d

Consider a circle of radius R, and suppose that a point within the circle is randomly

chosen in such a manner that all regions within the circle of equal area are equally

likely to contain the point. (In other words, the point is uniformly distributed within

the circle.) If we let the center of the circle denote the origin and deﬁne X and Y to be

the coordinates of the point chosen (Figure 6.1), then, since (X, Y) is equally likely

to be near each point in the circle, it follows that the joint density function of X and

Y is given by

f (x, y) =

c ifx

+ y

… R

0ifx

+ y

> R

for some value of c.

(a) Determine c.

(b) Find the marginal density functions of X and Y.

is less than or equal to a.

(d) Find E [D].

(0, 0)

(X, Y)

FIGURE 6.1: Joint probability distribution.

238 Chapter 6 Jointly Distributed Random Variables

Solution. (a) Because

−q

f (x, y) dy dx = 1

it follows that

…R

dy dx = 1

We can evaluate

…R

dy dx either by using polar coordinates or,

more simply, by noting that it represents the area of the circle and is thus equal

to πR

. Hence,

c =

πR

(b)

(x) =

−q

f (x, y) dy

πR

…R

πR

−c

dy, where c =

− x

πR

− x

… R

and it equals 0 when x

> R

. By symmetry, the marginal density of Y is given by

(y) =

πR

− y

… R

= 0 y

> R

+ Y

, the distance from the origin, is

obtained as follows: For 0 … a … R,

(a) = P{

+ Y

… a}

= P{X

+ Y

… a

}

…a

f (x, y) dy dx

πR

…a

dy dx

πa

πR

where we have used the fact that

…a

dy dx is the area of a circle of radius

a and thus is equal to π a

Section 6.1 Joint Distribution Functions 239

(d) From part (c), the density function of D is

(a) =

0 … a … R

Hence,

E[D] =

da =

EXAMPLE 1e

The joint density of X and Y is given by

f (x, y) =

−(x+y)

0 < x < q,0< y < q

0 otherwise

Find the density function of the random variable X/Y.

Solution. We start by computing the distribution function of X/Y.Fora > 0,

X/Y

(a) = P

… a

x/y…a

−(x+y)

dx dy

−(x+y)

dx dy

(1 − e

−ay

−y

−e

−y

−(a+1)y

a + 1

= 1 −

a + 1

Differentiation shows that the density function of X/Y is given by f

X/Y

(a) = 1/

(a + 1)

,0 < a < q. .

We can also deﬁne joint probability distributions for n random variables in exactly

thesamemanneraswedidforn = 2. For instance, the joint cumulative probabil-

ity distribution function F(a

, a

, ..., a

) of the n random variables X

, X

, ..., X

deﬁned by

F(a

, a

, ..., a

) = P{X

… a

, X

… a

, ..., X

… a

}

Further, the n random variables are said to be jointly continuous if there exists a

function f (x

, x

, ..., x

), called the joint probability density function, such that, for

any set C in n-space,

P{(X

, X

, ..., X

) ∈ C}=

···

,...,x

)∈C

f (x

, ..., x

)dx

···dx

240 Chapter 6 Jointly Distributed Random Variables

In particular, for any n sets of real numbers A

, A

, ..., A

P{X

∈ A

, X

, ∈ A

, ..., X

∈ A

}

n−1

···

f (x

, ..., x

) dx

···dx

EXAMPLE 1f The multinomial distribution

One of the most important joint distributions is the multinomial distribution, which

arises when a sequence of n independent and identical experiments is performed.

Suppose that each experiment can result in any one of r possible outcomes, with

respective probabilities p

, p

, ..., p



i=1

= 1. If we let X

denote the number of the

n experiments that result in outcome number i,then

P{X

= n

, X

= n

, ..., X

= n

! ···n

···p

(1.5)

whenever



i=1

= n.

Equation (1.5) is veriﬁed by noting that any sequence of outcomes for the n experi-

ments that leads to outcome i occurring n

times for i = 1, 2, ..., r will, by the assumed

independence of experiments, have probability p

...p

of occurring. Because

there are n!/(n

! ...n

!) such sequences of outcomes (there are n!/n

! ...n

!dif-

ferent permutations of n things of which n

are alike, n

are alike, ..., n

are alike),

Equation (1.5) is established. The joint distribution whose joint probability mass func-

tion is speciﬁed by Equation (1.5) is called the multinomial distribution. Note that

when r = 2, the multinomial reduces to the binomial distribution.

Note also that any sum of a ﬁxed set of the X



s will have a binomial distribution.

That is, if N ( {1, 2, ..., r},then



i∈N

will be a binomial random variable with

parameters n and p =



i∈N

. This follows because



i∈N

represents the number

of the n experiments whose outcome is in N, and each experiment will independently

have such an outcome with probability



i∈N

As an application of the multinomial distribution, suppose that a fair die is rolled

9 times. The probability that 1 appears three times, 2 and 3 twice each, 4 and 5 once

each, and 6 not at all is

3!2!2!1!1!0!

























3!2!2!





6.2 INDEPENDENT RANDOM VARIABLES

The random variables X and Y are said to be independent if, for any two sets of real

numbers A and B,

P{X ∈ A, Y ∈ B}=P{X ∈ A}P{Y ∈ B} (2.1)

In other words, X and Y are independent if, for all A and B, the events E

={X ∈ A}

and F

={Y ∈ B} are independent.

It can be shown by using the three axioms of probability that Equation (2.1) will

follow if and only if, for all a, b,

P{X … a, Y … b}=P{X … a}P{Y … b}

Section 6.2 Independent Random Variables 241

Hence, in terms of the joint distribution function F of X and Y, X and Y are inde-

pendent if

F(a, b) = F

(a)F

(b) for all a, b

When X and Y are discrete random variables, the condition of independence (2.1) is

equivalent to

p(x, y) = p

(x)p

(y) for all x, y (2.2)

The equivalence follows because, if Equation (2.1) is satisﬁed, then we obtain Equa-

tion (2.2) by letting A and B be, respectively, the one-point sets A ={x} and B ={y}.

Furthermore, if Equation (2.2) is valid, then, for any sets A, B,

P{X ∈ A, Y ∈ B}=



y∈B



x∈A

p(x, y)



y∈B



x∈A

(x)p

(y)



y∈B

(y)



x∈A

(x)

= P{Y ∈ B}P{X ∈ A}

and Equation (2.1) is established.

In the jointly continuous case, the condition of independence is equivalent to

f (x, y) = f

(x)f

(y) for all x, y

Thus, loosely speaking, X and Y are independent if knowing the value of one does

not change the distribution of the other. Random variables that are not independent

are said to be dependent.

EXAMPLE 2a

Suppose that n + m independent trials having a common probability of success p are

performed. If X is the number of successes in the ﬁrst n trials, and Y is the number

of successes in the ﬁnal m trials, then X and Y are independent, since knowing the

number of successes in the ﬁrst n trials does not affect the distribution of the number

of successes in the ﬁnal m trials (by the assumption of independent trials). In fact, for

integral x and y,

P{X = x, Y = y}=





(1 − p)

n−x





(1 − p)

m−y

0 … x … n,

0 … y … m

= P{X = x}P{Y = y}

In contrast, X and Z will be dependent, where Z is the total number of successes in

the n + m trials. (Why?) .

EXAMPLE 2b

Suppose that the number of people who enter a post ofﬁce on a given day is a Poisson

random variable with parameter λ. Show that if each person who enters the post ofﬁce

is a male with probability p and a female with probability 1 − p, then the number of

males and females entering the post ofﬁce are independent Poisson random variables

with respective parameters λp and λ(1 − p).

242 Chapter 6 Jointly Distributed Random Variables

Solution. Let X and Y denote, respectively, the number of males and females that

enter the post ofﬁce. We shall show the independence of X and Y by establishing

Equation (2.2). To obtain an expression for P{X = i, Y = j}, we condition on X + Y

as follows:

P{X = i, Y = j}=P{X = i, Y = j|X + Y = i + j}P{X + Y = i + j}

+ P{X = i, Y = j|X + Y Z i + j}P{X + Y Z i + j}

[Note that this equation is merely a special case of the formula P(E) = P

(E|F)P(F) +

P(E|F

)P(F

).]

Since P{X = i, Y = j|X + Y Z i + j} is clearly 0, we obtain

P{X = i, Y = j}=P{X = i, Y = j|X + Y = i + j}P{X + Y = i + j} (2.3)

Now, because X + Y is the total number of people who enter the post ofﬁce, it

follows, by assumption, that

P{X + Y = i + j}=e

−λ

i+j

(i + j )!

(2.4)

Furthermore, given that i + j people do enter the post ofﬁce, since each person

entering will be male with probability p, it follows that the probability that exactly

i of them will be male (and thus j of them female) is just the binomial probability



i + j



(1 − p)

. That is,

P{X = i, Y = j|X + Y = i + j}=



i + j



(1 − p)

(2.5)

Substituting Equations (2.4) and (2.5) into Equation (2.3) yields

P{X = i, Y = j}=



i + j



(1 − p)

−λ

i+j

(i + j )!

= e

−λ

(λp)

i!j!

[λ(1 − p)]

−λp

(λp)

−λ(1−p)

[λ(1 − p)]

(2.6)

Hence,

P{X = i}=e

−λp

(λp)



−λ(1−p)

[λ(1 − p)]

= e

−λp

(λp)

(2.7)

and similarly,

P{Y = j}=e

−λ(1−p)

[λ(1 − p)]

(2.8)

Equations (2.6), (2.7), and (2.8) establish the desired result. .

Section 6.2 Independent Random Variables 243

EXAMPLE 2c

A man and a woman decide to meet at a certain location. If each of them indepen-

dently arrives at a time uniformly distributed between 12 noon and 1

P.M., ﬁnd the

probability that the ﬁrst to arrive has to wait longer than 10 minutes.

Solution. If we let X and Y denote, respectively, the time past 12 that the man and

the woman arrive, then X and Y are independent random variables, each of which is

uniformly distributed over (0, 60). The desired probability, P{X + 10 < Y}+P{Y +

10 < X}, which, by symmetry, equals 2P{X + 10 < Y}, is obtained as follows:

2P{X + 10 < Y}=2

x+10<y

f (x, y) dx dy

= 2

x+10<y

(x)f

(y) dx dy

= 2

y−10





dx dy

(60)

(y − 10) dy

Our next example presents the oldest problem dealing with geometrical probabil-

ities. It was ﬁrst considered and solved by Buffon, a French naturalist of the 18th

century, and is usually referred to as Buffon’s needle problem.

EXAMPLE 2d Buffon’s needle problem

A table is ruled with equidistant parallel lines a distance D apart. A needle of length L,

where L … D, is randomly thrown on the table. What is the probability that the nee-

dle will intersect one of the lines (the other possibility being that the needle will be

completely contained in the strip between two lines)?

Solution. Let us determine the position of the needle by specifying (1) the distance X

from the middle point of the needle to the nearest parallel line and (2) the angle θ

between the needle and the projected line of length X. (See Figure 6.2.) The needle

will intersect a line if the hypotenuse of the right triangle in Figure 6.2 is less than

L/2—that is, if

cos θ

or X <

cos θ

␪

FIGURE 6.2

244 Chapter 6 Jointly Distributed Random Variables

As X varies between 0 and D/2andθ between 0 and π/2, it is reasonable to assume

that they are independent, uniformly distributed random variables over these respec-

tive ranges. Hence,

X <

cos θ

x<L/2 cos y

(x)f

(y) dx dy

πD

π/2

L/2 cos y

dx dy

πD

π/2

cos ydy

πD

∗

EXAMPLE 2e Characterization of the normal distribution

Let X and Y denote the horizontal and vertical miss distances when a bullet is ﬁred

at a target, and assume that

1. X and Y are independent continuous random variables having differentiable

density functions.

2. The joint density f (x, y) = f

(x)f

(y) of X and Y depends on (x, y) only through

+ y

Loosely put, assumption 2 states that the probability of the bullet landing on any

point of the x–y plane depends only on the distance of the point from the target and

not on its angle of orientation. An equivalent way of phrasing this assumption is to

say that the joint density function is rotation invariant.

It is a rather interesting fact that assumptions 1 and 2 imply that X and Y are

normally distributed random variables. To prove this, note ﬁrst that the assumptions

yield the relation

f (x, y) = f

(x)f

(y) = g(x

+ y

) (2.9)

for some function g. Differentiating Equation (2.9) with respect to x yields



(x)f

(y) = 2xg



+ y

) (2.10)

Dividing Equation (2.10) by Equation (2.9) gives



(x)

2xg



+ y

)

g(x

+ y

)



(x)

2xf

(x)



+ y

)

g(x

+ y

)

(2.11)

Because the value of the left-hand side of Equation (2.11) depends only on x,

whereas the value of the right-hand side depends on x

+ y

, it follows that the

left-hand side must be the same for all x. To see this, consider any x

, x

and let y

, y

be such that x

+ y

= x

+ y

. Then, from Equation (2.11), we obtain



)



+ y

)

g(x

+ y

)



+ y

)

g(x

+ y

)



)

Section 6.2 Independent Random Variables 245

Hence,



(x)

= c or

(log f

(x)) = cx

which implies, upon integration of both sides, that

log f

(x) = a +

or f

(x) = ke

Since

−q

(x) dx = 1, it follows that c is necessarily negative, and we may write

c =−1/σ

. Thus,

(x) = ke

−x

/2σ

That is, X is a normal random variable with parameters μ = 0andσ

. A similar

argument can be applied to f

(y) to show that

(y) =

√

2π σ

−y

/2σ

Furthermore, it follows from assumption 2 that σ

= σ

and that X and Y are thus

independent, identically distributed normal random variables with parameters μ = 0

and σ

. .

A necessary and sufﬁcient condition for the random variables X and Y to be

independent is for their joint probability density function (or joint probability mass

function in the discrete case) f (x, y) to factor into two terms, one depending only on

x and the other depending only on y.

Proposition 2.1. The continuous (discrete) random variables X and Y are indepen-

dent if and only if their joint probability density (mass) function can be expressed as

X,Y

(x, y) = h(x)g(y) − q < x < q, −q < y < q

Proof. Let us give the proof in the continuous case. First, note that independence

implies that the joint density is the product of the marginal densities of X and Y,so

the preceding factorization will hold when the random variables are independent.

Now, suppose that

X,Y

(x, y) = h(x)g(y)

Then

1 =

−q

X,Y

(x, y) dx dy

−q

h(x) dx

−q

g(y) dy

= C

where C

−q

h(x) dx and C

−q

g(y) dy. Also,

(x) =

−q

X,Y

(x, y) dy = C

h(x)

(y) =

−q

X,Y

(x, y) dx = C

g(y)

246 Chapter 6 Jointly Distributed Random Variables

Since C

= 1, it follows that

X,Y

(x, y) = f

(x)f

(y)

and the proof is complete.

EXAMPLE 2f

If the joint density function of X and Y is

f (x, y) = 6e

−2x

−3y

0 < x < q,0< y < q

and is equal to 0 outside this region, are the random variables independent? What if

the joint density function is

f (x, y) = 24xy 0 < x < 1, 0 < y < 1, 0 < x + y < 1

and is equal to 0 otherwise?

Solution. In the ﬁrst instance, the joint density function factors, and thus the random

variables, are independent (with one being exponential with rate 2 and the other

exponential with rate 3). In the second instance, because the region in which the joint

density is nonzero cannot be expressed in the form x ∈ A, y ∈ B, the joint density does

not factor, so the random variables are not independent. This can be seen clearly by

letting

I(x, y) =

1if0< x < 1, 0 < y < 1, 0 < x + y < 1

0 otherwise

and writing

f (x, y) = 24xy I(x, y)

which clearly does not factor into a part depending only on x and another depending

only on y. .

The concept of independence may, of course, be deﬁned for more than two random

variables. In general, the n random variables

, X

, ..., X

are said to be indepen-

dent if, for all sets of real numbers A

, A

, ..., A

P{X

∈ A

, X

∈ A

, ..., X

∈ A



i=1

P{X

∈ A

}

As before, it can be shown that this condition is equivalent to

P{X

… a

, X

… a

, ..., X

… a

}



i=1

P{X

… a

} for all a

, a

, ..., a

Finally, we say that an inﬁnite collection of random variables is independent if every

ﬁnite subcollection of them is independent.

EXAMPLE 2g How can a computer choose a random subset?

Most computers are able to generate the value of, or simulate, a uniform (0, 1) random

variable by means of a built-in subroutine that (to a high degree of approximation)

Section 6.2 Independent Random Variables 247

produces such “random numbers.” As a result, it is quite easy for a computer to sim-

ulate an indicator (that is, a Bernoulli) random variable. Suppose I is an indicator

variable such that

P{I = 1}=p = 1 − P{I = 0}

The computer can simulate I by choosing a uniform (0, 1) random number U and

then letting

I =

1ifU < p

0ifU Ú p

Suppose that we are interested in having the computer select k, k … n, of the num-

bers 1, 2, ..., n in such a way that each of the





subsets of size k is equally likely

to be chosen. We now present a method that will enable the computer to solve this

task. To generate such a subset, we will ﬁrst simulate, in sequence, n indicator vari-

ables I

, I

, ..., I

, of which exactly k will equal 1. Those i for which I

= 1 will then

constitute the desired subset.

To generate the random variables I

, ..., I

, start by simulating n independent uni-

form (0, 1) random variables U

, U

, ..., U

. Now deﬁne

⎧

⎪

⎨

⎪

⎩

1ifU

0 otherwise

and then, once I

, ..., I

are determined, recursively set

i+1

⎧

⎪

⎨

⎪

⎩

1ifU

i+1

k − (I

+ ··· + I

)

n − i

0 otherwise

In words, at the (i + 1)th stage we set I

i+1

equal to 1 (and thus put i + 1 into the

desired subset) with a probability equal to the remaining number of places in the sub-

set

⎛

⎝

namely, k −



j=1

⎞

⎠

, divided by the remaining number of possibilities (namely,

n − i). Hence, the joint distribution of I

, I

, ..., I

is determined from

P{I

= 1}=

P{I

i+1

= 1|I

, ..., I

k −



j=1

n − i

1 < i < n

The proof that the preceding formula results in all subsets of size k being equally

likely to be chosen is by induction on k + n. It is immediate when k + n = 2(that

is, when k = 1, n = 1), so assume it to be true whenever k + n … l. Now, suppose

that k + n = l + 1, and consider any subset of size k—say, i

… i

… ··· … i

—and

consider the following two cases.

248 Chapter 6 Jointly Distributed Random Variables

Case 1: i

= 1

P{I

= I

=···=I

= 1, I

= 0 otherwise}

= P{I

= 1}P{I

=···=I

= 1, I

= 0 otherwise|I

= 1}

Now given that I

= 1, the remaining elements of the subset are chosen as if a

subset of size k − 1 were to be chosen from the n − 1 elements 2, 3, ..., n. Hence,

by the induction hypothesis, the conditional probability that this will result in a given

subset of size k − 1 being selected is 1/



n − 1

k − 1



. Hence,

P{I

= I

=···=I

= 1, I

= 0 otherwise}



n − 1

k − 1







Case 2: i

Z 1

P{I

= I

=···=I

= 1, I

= 0 otherwise}

= P{I

=···=I

= 1, I

= 0 otherwise|I

= 0}P{I

= 0}



n − 1





1 −







where the induction hypothesis was used to evaluate the preceding conditional prob-

ability.

Thus, in all cases, the probability that a given subset of size k will be the subset

chosen is 1





. .

Remark. The foregoing method for generating a random subset has a very low

memory requirement. A faster algorithm that requires somewhat more memory is

presented in Section 10.1. (The latter algorithm uses the last k elements of a random

permutation of 1, 2, ..., n.) .

EXAMPLE 2h

Let X, Y, Z be independent and uniformly distributed over (0, 1). Compute P{X Ú

YZ}.

Solution. Since

X,Y,Z

(x, y, z) = f

(x)f

(y)f

(z) = 10… x … 1, 0 … y … 1, 0 … z … 1

we have

P{X Ú YZ}=

***

xÚyz

X, Y, Z

(x, y, z) dx dy dz

dx dy dz

Section 6.2 Independent Random Variables 249

(1 − yz) dy dz



1 −



EXAMPLE 2i Probabilistic interpretation of half-life

Let N(t) denote the number of nuclei contained in a radioactive mass of material at

time t. The concept of half-life is often deﬁned in a deterministic fashion by stating

this it is an empirical fact that, for some value h, called the half-life,

N(t) = 2

−t/h

N(0) t > 0

[Note that N(h) = N(0)/2.] Since the preceding implies that, for any nonnegative s

and t,

N(t + s) = 2

−(s+t)/h

N(0) = 2

−t/h

N(s)

it follows that no matter how much time s has already elapsed, in an additional time t

the number of existing nuclei will decrease by the factor 2

−t/h

Because the deterministic relationship just given results from observations of radio-

active masses containing huge numbers of nuclei, it would seem that it might be

consistent with a probabilistic interpretation. The clue to deriving the appropriate

probability model for half-life resides in the empirical observation that the propor-

tion of decay in any time interval depends neither on the total number of nuclei at

the beginning at the interval nor on the location of this interval (since N(t + s)/N(s)

depends neither on N(s) nor on s). Thus, it appears that the individual nuclei act inde-

pendently and with a memoryless life distribution. Consequently, since the unique life

distribution that is memoryless is the exponential distribution, and since exactly one-

half of a given amount of mass decays every h time units, we propose the following

probabilistic model for radioactive decay.

Probabilistic interpretation of the half-life h: The lifetimes of the individual nuclei

are independent random variables having a life distribution that is exponential with

median equal to h. That is, if L represents the lifetime of a given nucleus, then

P{L < t}=1 − 2

−t/h

(Because P{L < h}=

and the preceding can be written as

P{L < t}=1 − exp

−t

log 2

it can be seen that L indeed has an exponential distribution with median h.)

Note that, under the probabilistic interpretation of half-life just given, if one starts

with N(0) nuclei at time 0, then N(t), the number of nuclei that remain at time t,

will have a binomial distribution with parameters n = N(0) and p = 2

−t/h

. Results of

Chapter 8 will show that this interpretation of half-life is consistent with the determin-

istic model when considering the proportion of a large number of nuclei that decay

over a given time frame. However, the difference between the deterministic and prob-

abilistic interpretation becomes apparent when one considers the actual number of

250 Chapter 6 Jointly Distributed Random Variables

decayed nuclei. We will now indicate this with regard to the question of whether

protons decay.

There is some controversy over whether or not protons decay. Indeed, one theory

predicts that protons should decay with a half-life of about h = 10

years. To check

this prediction empirically, it has been suggested that one follow a large number of

protons for, say, one or two years and determine whether any of them decay within

that period. (Clearly, it would not be feasible to follow a mass of protons for 10

years to see whether one-half of it decays.) Let us suppose that we are able to keep

track of N(0) = 10

protons for c years. The number of decays predicted by the

deterministic model would then be given by

N(0) − N(c) = h(1 − 2

−c/h

)

1 − 2

−c/h

1/h

L lim

x→0

1 − 2

−cx

since

= 10

−30

L 0

= lim

x→0

(c2

−cx

log 2) by L’H

opital’s rule

= c log 2 L .6931c

For instance, the deterministic model predicts that in 2 years there should be 1.3863

decays, and it would thus appear to be a serious blow to the hypothesis that protons

decay with a half-life of 10

years if no decays are observed over those 2 years.

Let us now contrast the conclusions just drawn with those obtained from the prob-

abilistic model. Again, let us consider the hypothesis that the half-life of protons is

h = 10

years, and suppose that we follow h protons for c years. Since there is a

huge number of independent protons, each of which will have a very small probabil-

ity of decaying within this time period, it follows that the number of protons which

decay will have (to a very strong approximation) a Poisson distribution with parame-

ter equal to h(1 − 2

−c/h

) L c log 2. Thus,

P{0 decays}=e

−c log 2

= e

−log(2

)

and, in general,

P{n decays}=

−c

[c log 2]

n Ú 0

Thus we see that even though the average number of decays over 2 years is (as pre-

dicted by the deterministic model) 1.3863, there is 1 chance in 4 that there will not

be any decays, thereby indicating that such a result in no way invalidates the original

hypothesis of proton decay. .

Remark. Independence is a symmetric relation. The random variables X and Y

are independent if their joint density function (or mass function in the discrete case)

is the product of their individual density (or mass) functions. Therefore, to say that

X is independent of Y is equivalent to saying that Y is independent of X—or just

that X and Y are independent. As a result, in considering whether X is independent

of Y in situations where it is not at all intuitive that knowing the value of Y will not

change the probabilities concerning X, it can be beneﬁcial to interchange the roles of

Section 6.2 Independent Random Variables 251

X and Y and ask instead whether Y is independent of X. The next example illustrates

this point. .

EXAMPLE 2j

If the initial throw of the dice in the game of craps results in the sum of the dice

equaling 4, then the player will continue to throw the dice until the sum is either 4 or

7. If this sum is 4, then the player wins, and if it is 7, then the player loses. Let N denote

the number of throws needed until either 4 or 7 appears, and let X denote the value

(either 4 or 7) of the ﬁnal throw. Is N independent of X? That is, does knowing which

of 4 or 7 occurs ﬁrst affect the distribution of the number of throws needed until that

number appears? Most people do not ﬁnd the answer to this question to be intuitively

obvious. However, suppose that we turn it around and ask whether X is independent

of N. That is, does knowing how many throws it takes to obtain a sum of either 4 or

7 affect the probability that that sum is equal to 4? For instance, suppose we know

that it takes n throws of the dice to obtain a sum of either 4 or 7. Does this affect the

probability distribution of the ﬁnal sum? Clearly not, since all that is important is that

its value is either 4 or 7, and the fact that none of the ﬁrst n − 1 throws were either 4

or 7 does not change the probabilities for the nth throw. Thus, we can conclude that

X is independent of N, or equivalently, that N is independent of X.

As another example, let X

, X

, ... be a sequence of independent and identically

distributed continuous random variables, and suppose that we observe these random

variables in sequence. If X

> X

for each i = 1, ..., n − 1, then we say that X

a record value. That is, each random variable that is larger than all those preceding

it is called a record value. Let A

denote the event that X

is a record value. Is A

n+1

independent of A

? That is, does knowing that the nth random variable is the largest

of the ﬁrst n change the probability that the (n + 1)st random variable is the largest

of the ﬁrst n + 1? While it is true that A

n+1

is independent of A

, this may not be

intuitively obvious. However, if we turn the question around and ask whether A

independent of A

n+1

, then the result is more easily understood. For knowing that

the (n + 1)st value is larger than X

, ..., X

clearly gives us no information about

the relative size of X

among the ﬁrst n random variables. Indeed, by symmetry, it is

clear that each of these n random variables is equally likely to be the largest of this

set, so P(A

n+1

) = P(A

) = 1/n. Hence, we can conclude that A

and A

n+1

are

independent events. .

Remark. It follows from the identity

P{X

… a

, ..., X

… a

}

= P{X

… a

}P{X

… a

}···P{X

… a

, ..., X

n−1

… a

n−1

}

that the independence of X

, ..., X

can be established sequentially. That is, we can

show that these random variables are independent by showing that

is independent of X

, X

is independent of X

, X

is independent of X

, ..., X

n−1

252 Chapter 6 Jointly Distributed Random Variables

6.3 SUMS OF INDEPENDENT RANDOM VARIABLES

It is often important to be able to calculate the distribution of X + Y from the dis-

tributions of X and Y when X and Y are independent. Suppose that X and Y are

independent, continuous random variables having probability density functions f

and f

. The cumulative distribution function of X + Y is obtained as follows:

X+Y

(a) = P{X + Y … a}

x+y…a

(x)f

(y) dx dy

−q

a−y

−q

(x)f

(y) dx dy (3.1)

−q

a−y

−q

(x) dxf

(y) dy

−q

(a − y)f

(y) dy

The cumulative distribution function F

X+Y

is called the convolution of the distribu-

tions F

and F

(the cumulative distribution functions of X and Y, respectively).

By differentiating Equation (3.1), we ﬁnd that the probability density function

X+Y

of X + Y is given by

X+Y

(a) =

−q

(a − y)f

(y) dy

−q

(a − y)f

(y) dy (3.2)

−q

(a − y)f

(y) dy

6.3.1 Identically Distributed Uniform Random Variables

It is not difﬁcult to determine the density function of the sum of two independent

uniform (0, 1) random variables.

EXAMPLE 3a Sum of two independent uniform random variables

If X and Y are independent random variables, both uniformly distributed on (0, 1),

calculate the probability density of X + Y.

Solution. From Equation (3.2), since

(a) = f

(a) =

10< a < 1

0 otherwise

we obtain

X+Y

(a) =

(a − y) dy

For 0 … a … 1, this yields

X+Y

(a) =

dy = a

Section 6.3 Sums of Independent Random Variables 253

(

)

FIGURE 6.3: Triangular density function.

For 1 < a < 2, we get

X+Y

(a) =

a−1

dy = 2 − a

Hence,

X+Y

(a) =

⎧

⎨

⎩

a 0 … a … 1

2 − a 1 < a < 2

0 otherwise

Because of the shape of its density function (see Figure 6.3), the random variable

X + Y is said to have a triangular distribution. .

Now, suppose that X

, X

, ..., X

are independent uniform (0, 1) random variables,

and let

(x) = P{X

+ ... + X

… x}

Whereas a general formula for F

(x) is messy, it has a particularly nice form when

x … 1. Indeed, we now use mathematical induction to prove that

(x) = x

/n!, 0 … x … 1

Because the proceeding equation is true for n = 1, assume that

n−1

(x) = x

n−1

/(n − 1)!, 0 … x … 1

Now, writing



i=1

n−1



i=1

+ X

and using the fact that the X

are all nonnegative, we see from Equation 3.1 that, for

0 … x … 1,

(x) =

n−1

(x − y)f

(y)dy

(n − 1)!

(x − y)

n−1

dy by the induction hypothesis

= x

/n!

which completes the proof.

254 Chapter 6 Jointly Distributed Random Variables

For an interesting application of the preceding formula, let us use it to determine

the expected number of independent uniform (0, 1) random variables that need to

be summed to exceed 1. That is, with X

, X

, ... being independent uniform (0, 1)

random variables, we want to determine E[N], where

N = min{n : X

+ ... + X

> 1}

Noting that N is greater than n > 0 if and only if X

+ ... + X

… 1, we see that

P{N > n}=F

(1) = 1/n!, n > 0

Because

P{N > 0}=1 = 1/0!

we see that, for n > 0,

P{N = n}=P{N > n − 1}−P{N > n}=

(n − 1)!

−

n − 1

Therefore,

E[N] =



n=1

n(n − 1)



n=2

(n − 2)!

= e

That is, the mean number of independent uniform (0, 1) random variables that must

be summed for the sum to exceed 1 is equal to e.

6.3.2 Gamma Random Variables

Recall that a gamma random variable has a density of the form

f (y) =

λe

−λy

(λy)

t−1

(t)

0 < y < q

An important property of this family of distributions is that, for a ﬁxed value of λ,it

is closed under convolutions.

Proposition 3.1. If X and Y are independent gamma random variables with respec-

tive parameters (s, λ) and (t, λ),thenX + Y is a gamma random variable with param-

eters (s + t, λ).

Proof. Using Equation (3.2), we obtain

X+Y

(a) =

(s)(t)

λe

−λ(a−y)

[λ(a − y)]

s−1

λe

−λy

(λy)

t−1

= Ke

−λa

(a − y)

s−1

t−1

= Ke

−λa

s+t−1

(1 − x)

s−1

t−1

dx by letting x =

= Ce

−λa

s+t−1

Section 6.3 Sums of Independent Random Variables 255

where C is a constant that does not depend on a. But, as the preceding is a density

function and thus must integrate to 1, the value of C is determined, and we have

X+Y

(a) =

λe

−λa

(λa)

s+t−1

(s + t)

Hence, the result is proved.

It is now a simple matter to establish, by using Proposition 3.1 and induction, that if

, i = 1, ..., n are independent gamma random variables with respective parameters

, λ), i = 1, ..., n,then



i=1

is gamma with parameters





i=1

, λ



. We leave the

proof of this statement as an exercise.

EXAMPLE 3b

Let X

, X

, ..., X

be n independent exponential random variables, each having param-

eter λ. Then, since an exponential random variable with parameter λ is the same as

a gamma random variable with parameters (1, λ), it follows from Proposition 3.1 that

+ X

+ ··· + X

is a gamma random variable with parameters (n, λ). .

If Z

, Z

, ..., Z

are independent standard normal random variables, then Y K



i=1

is said to have the chi-squared (sometimes seen as χ

) distribution with n

degrees of freedom. Let us compute the density function of Y.Whenn = 1, Y = Z

and from Example 7b of Chapter 5, we see that its probability density function is

given by

(y) =

√

(

√

y) + f

(−

√

y)]

√

2π

−y/2

(y/2)

1/2−1

√

But we recognize the preceding as the gamma distribution with parameters





[A by-product of this analysis is that 





√

π.] But since each Z

is gamma





, it follows from Proposition 3.1 that the χ

distribution with n degrees of

freedom is just the gamma distribution with parameters



n/2,



and hence has a

probability density function given by

(y) =

−y/2





n/2−1







y > 0

−y/2

n/2−1

n/2







y > 0

256 Chapter 6 Jointly Distributed Random Variables

When n is an even integer, (n/2) = [(n/2) − 1]!, whereas when n is odd, (n/2) can

be obtained from iterating the relationship (t) = (t − 1)(t − 1) and then using

the previously obtained result that 





√

π. [For instance, 

















√

π.]

In practice, the chi-squared distribution often arises as the distribution of the square

of the error involved when one attempts to hit a target in n-dimensional space when

the coordinate errors are taken to be independent standard normal random variables.

It is also important in statistical analysis.

6.3.3 Normal Random Variables

We can also use Equation (3.2) to prove the following important result about normal

random variables.

Proposition 3.2. If X

, i = 1, ..., n, are independent random variables that are nor-

mally distributed with respective parameters μ

, σ

, i = 1, ..., n,then



i=1

is normally

distributed with parameters



i=1

and



i=1

Proof of Proposition 3.2: To begin, let X and Y be independent normal random

variables with X having mean 0 and variance σ

and Y having mean 0 and vari-

ance 1. We will determine the density function of X + Y by utilizing Equation (3.2).

Now, with

c =

2σ

1 + σ

2σ

we have

(a − y)f

(y) =

√

2πσ

exp

−

(a − y)

2σ

√

2π

exp

−

2πσ

exp

−

2σ

exp

−c



− 2y

1 + σ



Hence, from Equation (3.2),

X+Y

(a) =

2πσ

exp

−

2σ

exp

2σ

(1 + σ

)

−q

exp

−c



y −

1 + σ



2πσ

exp

−

2(1 + σ

)

−q

exp{−cx

}dx

= C exp

−

2(1 + σ

)

where C does not depend on a. But this implies that X + Y is normal with mean 0

and variance 1 + σ

Section 6.3 Sums of Independent Random Variables 257

Now, suppose that X

and X

are independent normal random variables with X

having mean μ

and variance σ

, i = 1, 2. Then

+ X

= σ



− μ



+ μ

But since (X

− μ

)/σ

is normal with mean 0 and variance σ

/σ

,and(X

− μ

)/σ

is normal with mean 0 and variance 1, it follows from our previous result that (X

−

)/σ

+ (X

− μ

)/σ

is normal with mean 0 and variance 1 + σ

/σ

, implying

that X

+ X

is normal with mean μ

+ μ

and variance σ

(1 + σ

/σ

) = σ

+ σ

Thus, Proposition 3.2 is established when n = 2. The general case now follows by

induction. That is, assume that Proposition 3.2 is true when there are n − 1 random

variables. Now consider the case of n, and write



i=1

n−1



i=1

+ X

By the induction hypothesis,

n−1



i=1

is normal with mean

n−1



i=1

and variance

n−1



i=1

Therefore, by the result for n = 2,



i=1

is normal with mean



i=1

and variance



i=1

EXAMPLE 3c

A basketball team will play a 44-game season. Twenty-six of these games are against

class A teams and 18 are against class B teams. Suppose that the team will win each

game against a class A team with probability .4 and will win each game against a class

B team with probability .7. Suppose also that the results of the different games are

independent. Approximate the probability that

(a) the team wins 25 games or more;

(b) the team wins more games against class A teams than it does against class B

teams.

Solution. (a) Let X

and X

respectively denote the number of games the team

wins against class A and against class B teams. Note that X

and X

are independent

binomial random variables and

E[X

] = 26(.4) = 10.4Var(X

) = 26(.4)(.6) = 6.24

E[X

] = 18(.7) = 12.6Var(X

) = 18(.7)(.3) = 3.78

By the normal approximation to the binomial, X

and X

will have approximately

the same distribution as would independent normal random variables with the pre-

ceding expected values and variances. Hence, by Proposition 3.2, X

+ X

will have

258 Chapter 6 Jointly Distributed Random Variables

approximately a normal distribution with mean 23 and variance 10.02. Therefore,

letting Z denote a standard normal random variable, we have

P{X

+ X

Ú 25}=P{X

+ X

Ú 24.5}

= P

+ X

− 23

√

10.02

24.5 − 23

√

10.02

L P

Z Ú

1.5

√

10.02

L 1 − P{Z <.4739}

L .3178

(b)WenotethatX

− X

will have approximately a normal distribution with

mean −2.2 and variance 10.02. Hence,

P{X

− X

Ú 1}=P{X

− X

Ú .5}

= P

− X

+ 2.2

√

10.02

.5 + 2.2

√

10.02

L P

Z Ú

2.7

√

10.02

L 1 − P{Z <.8530}

L .1968

Therefore, there is approximately a 31.78 percent chance that the team will win at

least 25 games and approximately a 19.68 percent chance that it will win more games

against class A teams than against class B teams. .

The random variable Y is said to be a lognormal random variable with parame-

ters μ and σ if log (Y) is a normal random variable with mean μ and variance σ

That is, Y is lognormal if it can be expressed as

Y = e

where X is a normal random variable.

EXAMPLE 3d

Starting at some ﬁxed time, let S(n) denote the price of a certain security at the end

of n additional weeks, n Ú 1. A popular model for the evolution of these prices

assumes that the price ratios S(n)/S(n − 1), n Ú 1, are independent and identi-

cally distributed lognormal random variables. Assuming this model, with parameters

μ = .0165, σ = .0730, what is the probability that

(a) the price of the security increases over each of the next two weeks?

(b) the price at the end of two weeks is higher than it is today?

Solution. Let Z be a standard normal random variable. To solve part (a), we use the

fact that log(x) increases in x to conclude that x > 1 if and only if log(x)>log(1) = 0.

As a result, we have

Section 6.3 Sums of Independent Random Variables 259

S(1)

S(0)

> 1

= P

log



S(1)

S(0)



> 0

= P

Z >

−.0165

.0730

= P{Z <.2260}

= .5894

In other words, the probability that the price is up after one week is .5894. Since the

successive price ratios are independent, the probability that the price increases over

each of the next two weeks is (.5894)

= .3474.

To solve part (b), we reason as follows:

S(2)

S(0)

> 1

= P

S(2)

S(1)

S(0)

> 1

= P

log



S(2)

S(1)



+ log



S(1)

S(0)



> 0

However, log



S(2)

S(1)



+ log



S(1)

S(0)



, being the sum of two independent normal random

variables with a common mean .0165 and a common standard deviation .0730, is a

normal random variable with mean .0330 and variance 2(.0730)

. Consequently,

S(2)

S(0)

> 1

= P

Z >

−.0330

.0730

√

= P{Z <.31965}

= .6254

6.3.4 Poisson and Binomial Random Variables

Rather than attempt to derive a general expression for the distribution of X + Y in

the discrete case, we shall consider some examples.

EXAMPLE 3e Sums of independent Poisson random variables

If X and Y are independent Poisson random variables with respective parameters λ

and λ

, compute the distribution of X + Y.

Solution. Because the event {X + Y = n}may be written as the union of the disjoint

events {X = k, Y = n − k},0 … k … n, we have

P{X + Y = n}=



k=0

P{X = k, Y = n − k}



k=0

P{X = k}P{Y = n − k}



k=0

−λ

n−k

(n − k)!

260 Chapter 6 Jointly Distributed Random Variables

= e

−(λ

+λ

)



k=0

n−k

k!(n − k)!

−(λ

+λ

)



k=0

k!(n − k)!

n−k

−(λ

+λ

)

(λ

+ λ

)

Thus, X

+ X

has a Poisson distribution with parameter λ

+ λ

. .

EXAMPLE 3f Sums of independent binomial random variables

Let X and Y be independent binomial random variables with respective parameters

(n, p)and(m, p). Calculate the distribution of X + Y.

Solution. Recalling the interpretation of a binomial random variable, and without

any computation at all, we can immediately conclude that X + Y is binomial with

parameters (n + m, p). This follows because X represents the number of successes in

n independent trials, each of which results in a success with probability p; similarly,

Y represents the number of successes in m independent trials, each of which results

in a success with probability p. Hence, given that X and Y are assumed independent,

it follows that X + Y represents the number of successes in n + m independent

trials when each trial has a probability p of resulting in a success. Therefore, X + Y

is a binomial random variable with parameters (n + m, p). To check this conclusion

analytically, note that

P{X + Y = k}=



i=0

P{X = i, Y = k − i}



i=0

P{X = i}P{Y = k − i}



i=0





n−i



k − i



k−i

m−k+i

where q = 1 − p and where





= 0 when j < 0. Thus,

P{X + Y = k}=p

n+m−k



i=0





k − i



and the conclusion follows upon application of the combinatorial identity



n + m





i=0





k − i



6.3.5 Geometric Random Variables

Let X

, ..., X

be independent geometric random variables, with X

having parame-

ter p

for i = 1, ..., n. We are interested in computing the probability mass function

Section 6.3 Sums of Independent Random Variables 261

of their sum S



i=1

. For an application, consider n coins, with coin i having

probability p

of coming up heads when ﬂipped, i = 1, ..., n. Suppose that coin 1 is

ﬂipped until heads appears, at which point coin 2 is ﬂipped until it shows heads, and

then coin 3 is ﬂipped until it shows heads, and so on. If we let X

denote the number

of ﬂips made with coin i,thenX

, X

, ..., X

will be independent geometric random

variables with respective parameters p

, p

, ..., p

,and S



i=1

will represent

the total number of ﬂips. If all the p

are equal—say, all p

= p—then S

has the same

distribution as the number of ﬂips of a coin having probability p of coming up heads

that are needed to obtain a total of n heads, and so S

is a negative binomial random

variable with probability mass function

P{S

= k}=



k − 1

n − 1



(1 − p)

k−n

, k Ú n

As a prelude to determining the probability mass function of S

when the p

are all

distinct, let us ﬁrst consider the case n = 2. Letting q

= 1 − p

, j = 1, 2, we obtain

P(S

= k) =

k−1



j=1

P{X

= j, X

= k − j}

k−1



j=1

P{X

= j}P{X

= k − j} (by independence)

k−1



j=1

j−1

k−j−1

= p

k−2

k−1



j=1

)

j−1

= p

k−2

1 − (q

)

k−1

1 − q

k−1

− q

−

k−1

− q

= p

k−1

− p

+ p

k−1

− p

If we now let n = 3 and compute P{S

= k} by starting with the identity

P{S

= k}=

k−1



j=1

P{S

= j, X

= k − j}=

k−1



j=1

P{S

= j}P{X

= k − j}

and then substituting the derived formula for the mass function of S

, we would

obtain, after some computations,

P{S

= k}=p

k−1

− p

+ p

k−1

− p

+ p

k−1

− p

262 Chapter 6 Jointly Distributed Random Variables

The mass functions of S

and S

lead to the following conjecture for the mass function

of S

Proposition 3.3. Let X

, ..., X

be independent geometric random variables, with X

having parameter p

for i = 1, ..., n. If all the p

are distinct, then, for k Ú n,

P{S

= k}=



i=1

k−1



jZi

− p

Proof of Proposition 3.3: We will prove this proposition by induction on the value

of n + k. Because the proposition is true when n = 2, k = 2, take as the induction

hypothesis that it is true for any k Ú n for which n + k … r. Now, suppose k Ú n are

such that n + k = r + 1. To compute P{S

= k}, we condition on whether X

= 1.

This gives

P{S

= k}=P{S

= k|X

= 1}P{X

= 1}+P{S

= k|X

> 1}P{X

> 1}

= P{S

= k|X

= 1}p

+ P{S

= k|X

> 1}q

Now,

P{S

= k|X

= 1}=P{S

n−1

= k − 1|X

= 1}

= P{S

n−1

= k − 1} (by independence)

n−1



i=1

k−2



iZj…n−1

− p

(by the induction hypothesis)

Now, if X is geometric with parameter p, then the conditional distribution of X given

that it is larger than 1 is the same as the distribution of 1 (the ﬁrst failed trial) plus

a geometric with parameter p (the number of additional trials after the ﬁrst until a

success occurs). Consequently,

P{S

= k|X

> 1}=P{X

+ ... + X

n−1

+ X

+ 1 = k}

= P{S

= k − 1}



i=1

k−2



iZj…n

− p

where the ﬁnal equality follows from the induction hypothesis. Thus, from the pre-

ceding, we obtain

P{S

= k}=p

n−1



i=1

k−2



iZj…n−1

− p

+ q



i=1

k−2



iZj…n

− p

= p

n−1



i=1

k−2



iZj…n−1

− p

+ q

n−1



i=1

k−2



iZj…n

− p

+ q

k−2



j<n

− p

n−1



i=1

k−2

(1 +

− p

)



iZj…n−1

− p

+ p

k−1



j<n

− p

Section 6.4 Conditional Distributions: Discrete Case 263

Now, using that

1 +

− p

+ q

− p

the preceding gives

P{S

= k}=

n−1



i=1

k−1



iZj…n

− p

+ p

k−1



j<n

− p



i=1

k−1



jZi

− p

and the proof by induction is complete. .

6.4 CONDITIONAL DISTRIBUTIONS: DISCRETE CASE

Recall that, for any two events E and F, the conditional probability of E given F is

deﬁned, provided that P(F)>0, by

P(E|F) =

P(EF)

P(F)

Hence, if X and Y are discrete random variables, it is natural to deﬁne the conditional

probability mass function of X given that Y = y,by

X|Y

(x|y) = P{X = x|Y = y}

P{X = x, Y = y}

P{Y = y}

p(x, y)

(y)

for all values of y such that p

(y)>0. Similarly, the conditional probability distribu-

tion function of X given that Y = y is deﬁned, for all y such that p

(y)>0, by

X|Y

(x|y) = P{X … x|Y = y}



a…x

X|Y

(a|y)

In other words, the deﬁnitions are exactly the same as in the unconditional case,

except that everything is now conditional on the event that Y = y.IfX is indepen-

dent of Y, then the conditional mass function and the distribution function are the

same as the respective unconditional ones. This follows because if X is independent

of Y,then

X|Y

(x|y) = P{X = x|Y = y}

P{X = x, Y = y}

P{Y = y}

P{X = x}P{Y = y}

P{Y = y}

= P{X = x}

264 Chapter 6 Jointly Distributed Random Variables

EXAMPLE 4a

Suppose that p(x, y), the joint probability mass function of X and Y, is given by

p(0, 0) = .4 p(0, 1) = .2 p(1, 0) = .1 p(1, 1) = .3

Calculate the conditional probability mass function of X given that Y = 1.

Solution. We ﬁrst note that

(1) =



p(x,1) = p(0, 1) + p(1, 1) = .5

Hence,

X|Y

(0|1) =

p(0, 1)

(1)

and

X|Y

(1|1) =

p(1, 1)

(1)

EXAMPLE 4b

If X and Y are independent Poisson random variables with respective parameters λ

and λ

, calculate the conditional distribution of X given that X + Y = n.

Solution. We calculate the conditional probability mass function of X given that X +

Y = n as follows:

P{X = k|X + Y = n}=

P{X = k, X + Y = n}

P{X + Y = n}

P{X = k, Y = n − k}

P{X + Y = n}

P{X = k}P{Y = n − k}

P{X + Y = n}

where the last equality follows from the assumed independence of X and Y. Recalling

(Example 3e) that X + Y has a Poisson distribution with parameter λ

+ λ

, we see

that the preceding equals

P{X = k|X + Y = n}=

−λ

n−k

(n − k)!



−(λ

+λ

)

(λ

+ λ

)



−1

(n − k)! k!

n−k

(λ

+ λ

)





+ λ





+ λ



n−k

In other words, the conditional distribution of X given that X + Y = n is the binomial

distribution with parameters n and λ

/(λ

+ λ

). .

We can also talk about joint conditional distributions, as is indicated in the next

two examples.

Section 6.4 Conditional Distributions: Discrete Case 265

EXAMPLE 4c

Consider the multinomial distribution with joint probability mass function

P{X

= n

, i = 1, ..., k}=

! ···n

···p

, n

Ú 0,



i=1

= n

Such a mass function results when n independent trials are performed, with each

trial resulting in outcome i with probability p



i=1

= 1. The random variables

, i = 1, ..., k, represent, respectively, the number of trials that result in outcome i,

i = 1, ..., k. Suppose we are given that n

of the trials resulted in outcome j,forj =

r + 1, ..., k, where



j=r+1

= m … n. Then, because each of the other n − m trials

must have resulted in one of the trials 1, ..., r, it would seem that the conditional dis-

tribution of X

, ..., X

is the multinomial distribution on n − m trials with respective

trial outcome probabilities

P{outcome i|outcome is not any of r + 1, ..., k}=

, i = 1, ..., r

where F



i=1

is the probability that a trial results in one of the outcomes

1, ..., r.

Solution. To verify this intuition, let n

, ..., n

,besuchthat



i=1

= n − m. Then

P{X

= n

, ..., X

= n

r+1

= n

r+1

, ...X

= n

}

P{X

= n

, ..., X

= n

}

P{X

r+1

= n

r+1

, ...X

= n

}

!···n

···p

r+1

···p

(n−m)!n

r+1

!···n

n−m

r+1

···p

where the probability in the denominator was obtained by regarding outcomes 1, ..., r

as a single outcome having probability F

, thus showing that the probability is a multi-

nomial probability on n trials with outcome probabilities F

, p

r+1

, ..., p

. Because



i=1

= n − m, the preceding can be written as

P{X

= n

, ..., X

= n

r+1

= n

r+1

, ...X

= n

}

(n − m)!

! ···n

(

)

···(

)

and our intuition is upheld. .

EXAMPLE 4d

Consider n independent trials, with each trial being a success with probability p.

Given a total of k successes, show that all possible orderings of the k successes and

n − k failures are equally likely.

Solution. We want to show that, given a total of k successes, each of the





possible

orderings of k successes and n − k failures is equally likely. Let X denote the number

of successes, and consider any ordering of k successes and n − k failures, say, o =

(s, s, f , f , ..., f ). Then

266 Chapter 6 Jointly Distributed Random Variables

P(o|X = k) =

P(o, X = k)

P(X = k)

P(o)

P(X = k)

(1 − p)

n−k





(1 − p)

n−k





6.5 CONDITIONAL DISTRIBUTIONS: CONTINUOUS CASE

If X and Y have a joint probability density function f (x, y), then the conditional prob-

ability density function of X given that Y = y is deﬁned, for all values of y such that

(y)>0, by

X|Y

(x|y) =

f (x, y)

(y)

To motivate this deﬁnition, multiply the left-hand side by dx and the right-hand side

by (dx dy)/dy to obtain

X|Y

(x|y) dx =

f (x, y) dx dy

(y) dy

P{x … X … x + dx, y … Y … y + dy}

P{y … Y … y + dy}

= P{x … X … x + dx|y … Y … y + dy}

In other words, for small values of dx and dy, f

X|Y

(x|y)dx represents the conditional

probability that X is between x and x + dx given that Y is between y and y + dy.

The use of conditional densities allows us to deﬁne conditional probabilities of

events associated with one random variable when we are given the value of a second

random variable. That is, if X and Y are jointly continuous, then, for any set A,

P{X ∈ A|Y = y}=

X|Y

(x|y) dx

In particular, by letting A = (−q, a], we can deﬁne the conditional cumulative distri-

bution function of X given that Y = y by

X|Y

(a|y) K P{X … a|Y = y}=

−q

X|Y

(x|y) dx

The reader should note that, by using the ideas presented in the preceding discussion,

we have been able to give workable expressions for conditional probabilities, even

though the event on which we are conditioning (namely, the event {Y = y})has

probability 0.

EXAMPLE 5a

The joint density of X and Y is given by

f (x, y) =

x(2 − x − y) 0 < x < 1, 0 < y < 1

0 otherwise

Compute the conditional density of X given that Y = y, where 0 < y < 1.

Section 6.5 Conditional Distributions: Continuous Case 267

Solution. For 0 < x < 1, 0 < y < 1, we have

X|Y

(x|y) =

f (x, y)

(y)

f (x, y)

−q

f (x, y) dx

x(2 − x − y)

x(2 − x − y) dx

x(2 − x − y)

− y/2

6x(2 − x − y)

4 − 3y

EXAMPLE 5b

Suppose that the joint density of X and Y is given by

f (x, y) =

⎧

⎪

⎨

⎪

⎩

−x/y

−y

0 < x < q,0 < y < q

0 otherwise

Find P{X > 1|Y = y}.

Solution. We ﬁrst obtain the conditional density of X given that Y = y.

X|Y

(x|y) =

f (x, y)

(y)

−x/y

−y

(1/y)e

−x/y

Hence,

P{X > 1|Y = y}=

−x/y

=−e

−x/y

= e

−1/y

If X and Y are independent continuous random variables, the conditional density

of X given that Y = y is just the unconditional density of X. This is so because, in the

independent case,

X|Y

(x|y) =

f (x, y)

(y)

(x)f

(y)

= f

(x)

We can also talk about conditional distributions when the random variables are nei-

ther jointly continuous nor jointly discrete. For example, suppose that X is a continu-

ous random variable having probability density function f and N is a discrete random

variable, and consider the conditional distribution of X given that N = n. Then

268 Chapter 6 Jointly Distributed Random Variables

P{x < X < x + dx|N = n}

P{N = n|x < X < x + dx}

P{N = n}

P{x < X < x + dx}

and letting dx approach 0 gives

lim

dx→0

P{x < X < x + dx |N = n}

P{N = n|X = x}

P{N = n}

f (x)

thus showing that the conditional density of X given that N = n is given by

X|N

(x|n) =

P{N = n|X = x}

P{N = n}

f (x)

EXAMPLE 5c The Bivariate Normal Distribution

One of the most important joint distributions is the bivariate normal distribution.

We say that the random variables X, Y have a bivariate normal distribution if, for

constants μ

, μ

, σ

> 0, σ

> 0, −1 <ρ<1, their joint density function is given,

for all −q < x, y < q,by

f (x, y) =

2πσ

1 − ρ

exp

⎧

⎨

⎩

−

2(1 − ρ

)





x − μ





y − μ



− 2ρ

(x − μ

)(y − μ

)

⎤

⎦

⎫

⎪

⎬

⎪

⎭

We now determine the conditional density of X given that Y = y. In doing so, we

will continually collect all factors that do not depend on x and represent them by the

constants C

. The ﬁnal constant will then be found by using that

−q

X|Y

(x|y) dx = 1.

We have

X|Y

(x|y) =

f (x, y)

(y)

= C

f (x, y)

= C

exp

⎧

⎨

⎩

−

2(1 − ρ

)





x − μ



− 2ρ

x(y − μ

)



⎫

⎬

⎭

= C

exp

⎧

⎪

⎨

⎪

⎩

−

2σ

(1 − ρ

)

⎡

⎣

− 2x



+ ρ

(y − μ

)



⎤

⎦

⎫

⎪

⎬

⎪

⎭

= C

exp

⎧

⎪

⎨

⎪

⎩

−

2σ

(1 − ρ

)

⎡

⎣

x −



+ ρ

(y − μ

)



⎤

⎦

⎫

⎪

⎬

⎪

⎭

Recognizing the preceding equation as a normal density, we can conclude that, given

Y = y, the random variable X is normally distributed with mean μ

+ ρ

(y − μ

)

and variance σ

(1 − ρ

). Also, because the joint density of Y, X is exactly the same as

that of X, Y, except that μ

, σ

are interchanged with μ

, σ

, it similarly follows that

Section 6.5 Conditional Distributions: Continuous Case 269

the conditional distribution of Y given X = x is the normal distribution with mean

+ ρ

(x − μ

) and variance σ

(1 − ρ

). It follows from these results that the

necessary and sufﬁcient condition for the bivariate normal random variables X and

Y to be independent is that ρ = 0 (a result that also follows directly from their joint

density, because it is only when ρ = 0 that the joint density factors into two terms,

one depending only on x and the other only on y).

With C =

2πσ

√

1−ρ

, the marginal density of X can be obtained from

(x) =

−q

f (x, y) dy

= C

−q

exp

⎧

⎪

⎨

⎪

⎩

−

2(1 − ρ

)

⎡

⎣



x − μ





y − μ



−2ρ

(x − μ

)(y − μ

)



⎫

⎬

⎭

Making the change of variables w =

y−μ

gives

(x) = Cσ

exp

−

2(1 − ρ

)



x − μ



−q

exp

−

2(1 − ρ

)



− 2ρ

x − μ



= Cσ

exp

−

2(1 − ρ

)



x − μ



(1 − ρ

)

−q

exp

−

2(1 − ρ

)



w − ρ

x − μ



Because

2π(1 − ρ

)

−q

exp

−

2(1 − ρ

)



w −

(x − μ

)



dw = 1

we see that

(x) = Cσ

2π(1 − ρ

) e

−(x−μ

)

/2σ

√

2πσ

−(x−μ

)

/2σ

That is, X is normal with mean μ

and variance σ

. Similarly, Y is normal with

mean μ

and variance σ

. .

EXAMPLE 5d

Consider n + m trials having a common probability of success. Suppose, however,

that this success probability is not ﬁxed in advance but is chosen from a uniform (0, 1)

270 Chapter 6 Jointly Distributed Random Variables

population. What is the conditional distribution of the success probability given that

the n + m trials result in n successes?

Solution. If we let X denote the probability that a given trial is a success, then X

is a uniform (0, 1) random variable. Also, given that X = x,then + m trials are

independent with common probability of success x,soN, the number of successes,

is a binomial random variable with parameters (n + m, x). Hence, the conditional

density of X given that N = n is

X|N

(x|n) =

P{N = n|X = x}f

(x)

P{N = n}



n + m



(1 − x)

P{N = n}

0 < x < 1

= cx

(1 − x)

where c does not depend on x. Thus, the conditional density is that of a beta random

variable with parameters n + 1, m + 1.

The preceding result is quite interesting, for it states that if the original or prior

(to the collection of data) distribution of a trial success probability is uniformly dis-

tributed over (0, 1) [or, equivalently, is beta with parameters (1, 1)] then the posterior

(or conditional) distribution given a total of n successes in n + m trials is beta with

parameters (1 + n,1 + m). This is valuable, for it enhances our intuition as to what

it means to assume that a random variable has a beta distribution. .

∗

6.6 ORDER STATISTICS

Let X

, X

, ..., X

be n independent and identically distributed continuous random

variables having a common density f and distribution function F. Deﬁne

(1)

= smallest of X

, X

, ..., X

(2)

= second smallest of X

, X

, ..., X

(j)

= jth smallest of X

, X

, ..., X

(n)

= largest of X

, X

, ..., X

The ordered values X

(1)

… X

(2)

… ··· … X

(n)

are known as the order statistics cor-

responding to the random variables X

, X

, ..., X

. In other words, X

(1)

, ..., X

(n)

are

the ordered values of X

, ..., X

The joint density function of the order statistics is obtained by noting that the order

statistics X

(1)

, ..., X

(n)

will take on the values x

… x

… ··· … x

if and only if, for

some permutation (i

, i

, ..., i

) of (1, 2, ..., n),

= x

, X

= x

, ..., X

= x

Section 6.6 Order Statistics 271

Since, for any permutation (i

, ..., i

) of (1, 2, ..., n),

−

< X

< x

, ..., x

−

< X

< x

L ε

, ···, X

, ..., x

)

= ε

f (x

) ···f (x

)

= ε

f (x

) ···f (x

)

it follows that, for x

< x

< ··· < x

−

< X

(1)

< x

, ..., x

−

< X

(n)

< x

L n! ε

f (x

) ···f (x

)

Dividing by ε

and letting ε→0 yields

(1)

,...,X

(n)

, x

, ..., x

) = n!f(x

) ···f (x

) x

< x

< ··· < x

(6.1)

Equation (6.1) is most simply explained by arguing that, in order for the vector

X

(1)

, ..., X

(n)

 to equal x

, ..., x

, it is necessary and sufﬁcient for X

, ..., X

 to

equal one of the n! permutations of x

, ..., x

. Since the probability (density) that

X

, ..., X

 equals any given permutation of x

, ..., x

 is just f (x

) ···f (x

), Equa-

tion (6.1) follows.

EXAMPLE 6a

Along a road 1 mile long are 3 people “distributed at random.” Find the probability

that no 2 people are less than a distance of d miles apart when d …

Solution. Let us assume that “distributed at random” means that the positions of the

3 people are independent and uniformly distributed over the road. If X

denotes the

position of the ith person, then the desired probability is P{X

(i)

> X

(i−1)

+ d, i =

2, 3}. Because

(1)

(2)

(3)

, x

) = 3! 0 < x

< x

< 1

it follows that

P{X

(i)

> X

(i−1)

+ d, i = 2, 3}=

***

j−1

(1)

(2)

(3)

, x

) dx

= 3!

1−2d

1−d

= 6

1−2d

1−d

(1 − d − x

) dx

= 6

1−2d

1−2d−x

272 Chapter 6 Jointly Distributed Random Variables

where we have made the change of variables y

= 1 − d − x

. Continuing the string

of equalities yields

= 3

1−2d

(1 − 2d − x

)

= 3

1−2d

= (1 − 2d)

Hence, the desired probability that no 2 people are within a distance d of each other

when 3 people are uniformly and independently distributed over an interval of size

1is(1 − 2d)

when d …

. In fact, the same method can be used to prove that

when n people are distributed at random over the unit interval, the desired

probability is

[1 − (n − 1)d]

when d …

n − 1

The proof is left as an exercise. .

The density function of the jth-order statistic X

(j)

can be obtained either by inte-

grating the joint density function (6.1) or by direct reasoning as follows: In order for

(j)

to equal x, it is necessary for j − 1ofthen values X

, ..., X

to be less than

x, n − j of them to be greater than x, and 1 of them to equal x. Now, the probability

density that any given set of j − 1oftheX

’s are less than x, another given set of

n − j are all greater than x, and the remaining value is equal to x equals

[F(x)]

j−1

[1 − F(x)]

n−j

f (x)

Hence, since there are



j − 1, n − j,1



(n − j)!(j − 1)!

different partitions of the n random variables X

, ..., X

into the preceding three

groups, it follows that the density function of X

(j)

is given by

(j)

(x) =

(n − j )!(j − 1)!

[F(x)]

j−1

[1 − F(x)]

n−j

f (x) (6.2)

EXAMPLE 6b

When a sample of 2n + 1 random variables (that is, when 2n + 1 independent and

identically distributed random variables) is observed, the (n + 1)st smallest is called

the sample median. If a sample of size 3 from a uniform distribution over (0, 1) is

observed, ﬁnd the probability that the sample median is between

and

Solution. From Equation (6.2), the density of X

(2)

is given by

(2)

(x) =

1!1!

x(1 − x) 0 < x < 1

Section 6.6 Order Statistics 273

Hence,

< X

(2)

= 6

3/4

1/4

x(1 − x) dx

= 6

−

x=3/4

x=1/4

The cumulative distribution function of X

(j)

can be found by integrating Equa-

tion (6.2). That is,

(j)

(y) =

(n − j)!(j − 1)!

−q

[F(x)]

j−1

[1 − F(x)]

n−j

f (x) dx (6.3)

However, F

(j)

(y) could also have been derived directly by noting that the jth order

statistic is less than or equal to y if and only if there are j or more of the X

’s that are

less than or equal to y. Thus, because the number of X

’s that are less than or equal

to y is a binomial random variable with parameters n, p = F(y), it follows that

(j)

(y) = P{X

(j)

… y}=P{j or more of the X

’s are … y}



k=j





[F(y)]

[1 − F(y)]

n−k

(6.4)

If, in Equations (6.3) and (6.4), we take F to be the uniform (0, 1) distribution [that

is, f (x) = 1, 0 < x < 1], then we obtain the interesting analytical identity



k=j





(1 − y)

n−k

(n − j)!(j − 1)!

j−1

(1 − x)

n−j

dx 0 … y … 1 (6.5)

By employing the same type of argument that we used in establishing

Equation (6.2), we can show that the joint density function of the order statistics X

(i)

and X

(j)

when i < j is

(i)

(j)

, x

) =

(i − 1)!(j − i − 1)!(n − j)!

[F(x

)]

i−1

(6.6)

* [F(x

) − F(x

)]

j−i−1

[1 − F(x

)]

n−j

f (x

)f (x

)

for all x

< x

EXAMPLE 6c Distribution of the range of a random sample

Suppose that n independent and identically distributed random variables X

, X

, ...,

are observed. The random variable R deﬁned by R = X

(n)

− X

(1)

is called the

range of the observed random variables. If the random variables X

have distribution

function F and density function f , then the distribution of R can be obtained from

Equation (6.6) as follows: For a Ú 0,

274 Chapter 6 Jointly Distributed Random Variables

P{R … a}=P{X

(n)

− X

(1)

… a}

−x

…a

(1)

(n)

, x

) dx

−q

(n − 2)!

[F(x

) − F(x

)]

n−2

f (x

)f (x

) dx

Making the change of variable y = F(x

) − F(x

), dy = f (x

) dx

yields

[F(x

) − F(x

)]

n−2

f (x

) dx

F(x

+a)−F(x

)

n−2

n − 1

[F(x

+ a) − F(x

)]

n−1

Thus,

P{R … a}=n

−q

[F(x

+ a) − F(x

)]

n−1

f (x

) dx

(6.7)

Equation (6.7) can be evaluated explicitly only in a few special cases. One such case

is when the X

’s are all uniformly distributed on (0, 1). In this case, we obtain, from

Equation (6.7), that for 0 < a < 1,

P{R < a}=n

[F(x

+ a) − F(x

)]

n−1

f (x

) dx

= n

1−a

n−1

+ n

1−a

(1 − x

)

n−1

= n(1 − a)a

n−1

+ a

Differentiation yields the density function of the range: given in this case by

(a) =

n(n − 1)a

n−2

(1 − a) 0 … a … 1

0 otherwise

That is, the range of n independent uniform (0, 1) random variables is a beta random

variable with parameters n − 1, 2. .

6.7 JOINT PROBABILITY DISTRIBUTION OF FUNCTIONS OF RANDOM VARIABLES

Let X

and X

be jointly continuous random variables with joint probability density

function f

. It is sometimes necessary to obtain the joint distribution of the ran-

dom variables Y

and Y

, which arise as functions of X

and X

. Speciﬁcally, suppose

that Y

= g

, X

) and Y

= g

, X

) for some functions g

and g

Assume that the functions g

and g

satisfy the following conditions:

1. The equations y

= g

, x

) and y

= g

, x

) can be uniquely solved for x

and x

in terms of y

and y

, with solutions given by, say, x

= h

, y

), x

, y

2. The functions g

and g

have continuous partial derivatives at all points (x

, x

)

and are such that the 2 * 2 determinant

Section 6.7 Joint Distribution of Functions of Random Variables 275

J(x

, x

) =

∂g

∂x

∂g

∂x

∂g

∂x

∂g

∂x

∂g

∂x

∂g

∂x

−

∂g

∂x

∂g

∂x

Z 0

at all points (x

, x

Under these two conditions, it can be shown that the random variables Y

and Y

are jointly continuous with joint density function given by

, y

) = f

, x

)|J(x

, x

−1

(7.1)

where x

= h

, y

), x

= h

, y

A proof of Equation (7.1) would proceed along the following lines:

P{Y

… y

, Y

… y

, x

) :

, x

) … y

, x

) … y

, x

) dx

(7.2)

The joint density function can now be obtained by differentiating Equation (7.2) with

respect to y

and y

. That the result of this differentiation will be equal to the right-

hand side of Equation (7.1) is an exercise in advanced calculus whose proof will not

be presented in this book.

EXAMPLE 7a

Let X

and X

be jointly continuous random variables with probability density func-

tion f

.LetY

= X

+ X

, Y

= X

− X

. Find the joint density function of Y

and Y

in terms of f

Solution. Let g

, x

) = x

+ x

and g

, x

) = x

− x

. Then

J(x

, x

) =

1 −1

=−2

Also, since the equations y

= x

+ x

and y

= x

− x

have x

= (y

+ y

)/2, x

− y

)/2 as their solution, it follows from Equation (7.1) that the desired density is

, y

) =



+ y

− y



For instance, if X

and X

are independent uniform (0, 1) random variables, then

, y

) =

0 … y

+ y

… 2, 0 … y

− y

… 2

0 otherwise

276 Chapter 6 Jointly Distributed Random Variables

or if X

and X

are independent exponential random variables with respective param-

eters λ

and λ

,then

, y

)

⎧

⎪

⎨

⎪

⎩

exp

−λ



+ y



− λ



− y



+ y

Ú 0, y

− y

Ú 0

0 otherwise

⍜

FIGURE 6.4: •=Random point. (X, Y) = (R, Θ).

Finally, if X

and X

are independent standard normal random variables, then

, y

) =

4π

−[(y

)

/8+(y

−y

)

/8]

4π

−(y

)/4

√

4π

−y

√

4π

−y

Thus, not only do we obtain (in agreement with Proposition 3.2) that both X

+ X

and X

− X

are normal with mean 0 and variance 2, but we also conclude that these

two random variables are independent. (In fact, it can be shown that if X

and X

are independent random variables having a common distribution function F,then

+ X

will be independent of X

− X

if and only if F is a normal distribution

function.) .

EXAMPLE 7b

Let (X, Y) denote a random point in the plane, and assume that the rectangular

coordinates X and Y are independent standard normal random variables. We are

interested in the joint distribution of R, Θ, the polar coordinate representation of

(x, y). (See Figure 6.4.)

Suppose ﬁrst that X and Y are both positive. For x and y positive, letting r =

(x, y) =

+ y

and θ = g

(x, y) = tan

−1

y/x, we see that

∂g

∂x

+ y

∂g

∂y

+ y

Section 6.7 Joint Distribution of Functions of Random Variables 277

∂g

∂x

1 + (y/x)



−y



−y

+ y

∂g

∂y

x[1 + (y/x)

]

+ y

Hence,

J(x, y) =

+ y

)

3/2

+ y

)

3/2

+ y

Because the conditional joint density function of X, Y given that they are both

positive is

f (x, y|X > 0, Y > 0) =

f (x, y)

P(X > 0, Y > 0)

−(x

)/2

, x > 0, y > 0

we see that the conditional joint density function of R =

+ Y

and  =

tan

−1

(Y/X), given that X and Y are both positive, is

f (r, θ |X > 0, Y > 0) =

−r

2/2

,0<θ<π/2, 0 < r < q

Similarly, we can show that

f (r, θ |X < 0, Y > 0) =

−r

2/2

, π/2 <θ <π,0< r < q

f (r, θ |X < 0, Y < 0) =

−r

2/2

, π<θ<3π/2, 0 < r < q

f (r, θ |X > 0, Y < 0) =

−r

2/2

,3π/2 <θ <2π,0< r < q

As the joint density is an equally weighted average of these 4 conditional joint densi-

ties, we obtain that the joint density of R, Θ is given by

f (r, θ) =

2π

−r

0 <θ<2π ,0< r < q

Now, this joint density factors into the marginal densities for R and Θ,soR and Θ

are independent random variables, with Θ being uniformly distributed over (0, 2π)

and R having the Rayleigh distribution with density

f (r) = re

−r

0 < r < q

(For instance, when one is aiming at a target in the plane, if the horizontal and vertical

miss distances are independent standard normals, then the absolute value of the error

has the preceding Rayleigh distribution.)

This result is quite interesting, for it certainly is not evident a priori that a ran-

dom vector whose coordinates are independent standard normal random variables

will have an angle of orientation that not only is uniformly distributed, but also is

independent of the vector’s distance from the origin.

278 Chapter 6 Jointly Distributed Random Variables

If we wanted the joint distribution of R

and Θ, then, since the transformation

d = g

(x, y) = x

+ y

and θ = g

(x, y) = tan

−1

y/x has the Jacobian

J =

2x 2y

−y

+ y

= 2

it follows that

f (d, θ) =

−d/2

2π

0 < d < q,0<θ <2π

Therefore, R

and Θ are independent, with R

having an exponential distribution

with parameter

. But because R

= X

+ Y

, it follows by deﬁnition that R

has

a chi-squared distribution with 2 degrees of freedom. Hence, we have a veriﬁcation

of the result that the exponential distribution with parameter

is the same as the

chi-squared distribution with 2 degrees of freedom.

The preceding result can be used to simulate (or generate) normal random vari-

ables by making a suitable transformation on uniform random variables. Let U

and

be independent random variables, each uniformly distributed over (0, 1). We will

transform U

, U

into two independent unit normal random variables X

and X

by ﬁrst considering the polar coordinate representation (R, Θ) of the random vec-

tor (X

, X

). From the preceding, R

and Θ will be independent, and, in addition,

= X

+ X

will have an exponential distribution with parameter λ =

. But

−2logU

has such a distribution, since, for x > 0,

P{−2logU

< x}=P

log U

> −

= P{U

> e

−x/2

}

= 1 − e

−x/2

Also, because 2πU

is a uniform (0, 2π) random variable, we can use it to generate Θ.

That is, if we let

=−2logU

Θ = 2πU

then R

can be taken to be the square of the distance from the origin and θ can be

taken to be the angle of orientation of (X

, X

). Now, since X

= R cos Θ, X

R sin Θ, it follows that

−2logU

cos(2πU

)

−2logU

sin(2πU

)

are independent standard normal random variables. .

EXAMPLE 7c

If X and Y are independent gamma random variables with parameters (α, λ) and

(β, λ), respectively, compute the joint density of U = X + Y and V = X/(X + Y).

Section 6.7 Joint Distribution of Functions of Random Variables 279

Solution. The joint density of X and Y is given by

X,Y

(x, y) =

λe

−λx

(λx)

α−1

(α)

λe

−λy

(λy)

β−1

(β)

α+β

(α)(β)

−λ(x+y)

α−1

β−1

Now, if g

(x, y) = x + y, g

(x, y) = x/(x + y),then

∂g

∂x

∂g

∂y

= 1

∂g

∂x

(x + y)

∂g

∂y

=−

(x + y)

J(x, y) =

(x + y)

−x

(x + y)

=−

x + y

Finally, as the equations u = x + y, v = x/(x + y) have as their solutions x = uv, y =

u(1 − v), we see that

U,V

(u, v) = f

X,Y

[uv, u(1 − v)]u

λe

−λu

(λu)

α+β−1

(α + β)

α−1

(1 − v)

β−1

(α + β)

(α)(β)

Hence, X + Y and X/(X + Y) are independent, with X + Y having a gamma dis-

tribution with parameters (α + β, λ) and X/(X + Y) having a beta distribution with

parameters (α, β). The preceding reasoning also shows that B(α, β), the normalizing

factor in the beta density, is such that

B(α, β) K

α−1

(1 − v)

β−1

(α)(β)

(α + β)

This entire result is quite interesting. For suppose there are n + m jobs to be per-

formed, each (independently) taking an exponential amount of time with rate λ to be

completed and suppose that we have two workers to perform these jobs. Worker I

will do jobs 1, 2, ..., n, and worker II will do the remaining m jobs. If we let X and Y

denote the total working times of workers I and II, respectively, then (either from the

foregoing result or from Example 3b) X and Y will be independent gamma random

variables having parameters (n, λ) and (m, λ), respectively. It then follows that, inde-

pendently of the working time needed to complete all n + m jobs (that is, of X + Y),

the proportion of this work that will be performed by worker I has a beta distribution

with parameters (n, m). .

When the joint density function of the n random variables X

, X

, ..., X

is given

and we want to compute the joint density function of Y

, Y

, ..., Y

, where

= g

, ..., X

) Y

= g

, ..., X

), ... Y

= g

, ..., X

)

280 Chapter 6 Jointly Distributed Random Variables

the approach is the same—namely, we assume that the functions g

have continuous

partial derivatives and that the Jacobian determinant.

J(x

, ..., x

) =

∂g

∂x

∂g

∂x

···

∂g

∂x

∂g

∂x

∂g

∂x

···

∂g

∂x

∂g

∂x

∂g

∂x

···

∂g

∂x

Z 0

at all points (x

, ..., x

). Furthermore, we suppose that the equations y

, ..., x

), y

= g

, ..., x

), ..., y

= g

, ..., x

) have a unique solution, say,

= h

, ..., y

), ..., x

= h

, ..., y

). Under these assumptions, the joint den-

sity function of the random variables Y

is given by

,...,Y

, ..., y

) = f

,...,X

, ..., x

)|J(x

, ..., x

−1

(7.3)

where x

= h

, ..., y

), i = 1, 2, ..., n.

EXAMPLE 7d

Let X

, X

,andX

be independent standard normal random variables. If Y

= X

+ X

, Y

= X

− X

,andY

= X

− X

, compute the joint density function of

, Y

Solution. Letting Y

= X

+ X

, Y

= X

− X

, Y

= X

− X

, the Jacobian

of these transformations is given by

J =

111

1 −10

10−1

= 3

As the preceding transformations yield that

+ Y

− 2Y

+ Y

− 2Y

we see from Equation (7.3) that

, Y

, y

)

, X



+ y

− 2y

+ y

− 2y



Hence, as

, X

, x

) =

(2π)

3/2

−



i=1

we see that

, Y

, y

) =

3(2π)

3/2

−Q(y

, y

)/2

Section 6.7 Joint Distribution of Functions of Random Variables 281

where

Q(y

, y

)



+ y





− 2y

+ y





+ y

− 2y



−

EXAMPLE 7e

Let X

, X

, ..., X

be independent and identically distributed exponential random

variables with rate λ.Let

= X

+ ··· + X

i = 1, ..., n

(a) Find the joint density function of Y

, ..., Y

(b) Use the result of part (a) to ﬁnd the density of Y

Solution. (a) The Jacobian of the transformations Y

= X

, Y

= X

+ X

, ...,

= X

+ ··· + X

J =

1000··· 0

1100··· 0

1110··· 0

··· ···

1111··· 1

Since only the ﬁrst term of the determinant will be nonzero, we have J = 1.

Now, the joint density function of X

, ..., X

is given by

,...,X

, ..., x

) =



i=1

λe

−λx

0 < x

< q, i = 1, ..., n

Hence, because the preceding transformations yield

= Y

, X

= Y

− Y

, ..., X

= Y

− Y

i−1

, ..., X

= Y

− Y

n−1

it follows from Equation (7.3) that the joint density function of Y

, ..., Y

,...,Y

, y

, ..., y

)

= f

,...,X

, y

− y

, ..., y

− y

i−1

, ..., y

− y

n−1

)

= λ

exp

⎧

⎪

⎨

⎪

⎩

−λ

⎡

⎣



i=2

− y

i−1

)

⎤

⎦

⎫

⎪

⎬

⎪

⎭

= λ

−λy

0 < y

,0 < y

− y

i−1

, i = 2, ..., n

= λ

−λy

0 < y

< y

< ··· < y

282 Chapter 6 Jointly Distributed Random Variables

(b) To obtain the marginal density of Y

, let us integrate out the other variables one

at a time. Doing this gives

,...,Y

, ..., y

) =

−λy

= λ

−λy

0 < y

< y

< ··· < y

Continuing, we obtain

,...,Y

, ..., y

) =

−λy

= λ

−λy

0 < y

< y

< ··· < y

The next integration yields

,...,Y

, ..., y

) = λ

−λy

0 < y

< ··· < y

Continuing in this fashion gives

) = λ

n−1

(n − 1)!

−λy

0 < y

which, in agreement with the result obtained in Example 3b, shows that X

··· + X

is a gamma random variable with parameters n and λ.

∗

6.8 EXCHANGEABLE RANDOM VARIABLES

The random variables X

, X

, ..., X

are said to be exchangeable if, for every permu-

tation i

, ..., i

of the integers 1, ..., n,

P{X

… x

, X

… x

, ..., X

… x

}=P{X

… x

, X

… x

, ..., X

… x

}

for all x

, ..., x

. That is, the n random variables are exchangeable if their joint distri-

bution is the same no matter in which order the variables are observed.

Discrete random variables will be exchangeable if

P{X

= x

, X

= x

, ..., X

= x

}=P{X

= x

, X

= x

, ..., X

= x

}

for all permutations i

, ..., i

, and all values x

, ..., x

. This is equivalent to stating

that p(x

, x

, ..., x

) = P{X

= x

, ..., X

= x

} is a symmetric function of the vector

, ..., x

), which means that its value does not change when the values of the vector

are permuted.

EXAMPLE 8a

Suppose that balls are withdrawn one at a time and without replacement from an

urn that initially contains n balls, of which k are considered special, in such a manner

that each withdrawal is equally likely to be any of the balls that remain in the urn

at the time. Let X

= 1iftheith ball withdrawn is special and let X

= 0 otherwise.

We will show that the random variables X

, ..., X

are exchangeable. To do so, let

, ..., x

) be a vector consisting of k ones and n − k zeros. However, before consid-

ering the joint mass function evaluated at (x

, ..., x

), let us try to gain some insight by

Section 6.8 Exchangeable Random Variables 283

considering a ﬁxed such vector—for instance, consider the vector (1, 1, 0, 1, 0, ...,0,1),

which is assumed to have k ones and n − k zeros. Then

p(1, 1, 0, 1, 0, ...,0,1) =

k − 1

n − 1

n − k

n − 2

k − 2

n − 3

n − k − 1

n − 4

···

which follows because the probability that the ﬁrst ball is special is k/n, the condi-

tional probability that the next one is special is (k − 1)/(n − 1), the conditional

probability that the next one is not special is (n − k)/(n − 2), and so on. By the same

argument, it follows that p(x

, ..., x

) can be expressed as the product of n fractions.

The successive denominator terms of these fractions will go from n down to 1. The

numerator term at the location where the vector (x

, ..., x

) is 1 for the ith time is

k − (i − 1), and where it is 0 for the ith time it is n − k − (i − 1). Hence, since the

vector (x

, ..., x

) consists of k ones and n − k zeros, we obtain

p(x

, ..., x

) =

k!(n − k)!

= 0, 1,



i=1

= k

Since this is a symmetric function of (x

, ..., x

), it follows that the random variables

are exchangeable. .

Remark. Another way to obtain the preceding formula for the joint probability

mass function is to regard all the n balls as distinguishable from each other. Then,

since the outcome of the experiment is an ordering of these balls, it follows that

there are n! equally likely outcomes. Finally, because the number of outcomes hav-

ing special and nonspecial balls in speciﬁed places is equal to the number of ways of

permuting the special and the nonspecial balls among themselves, namely k!(n − k)!,

we obtain the preceding density function. .

It is easily seen that if X

, X

, ..., X

are exchangeable, then each X

has the same

probability distribution. For instance, if X and Y are exchangeable discrete random

variables, then

P{X = x}=



P{X = x, Y = y}=



P{X = y, Y = x}=P{Y = x}

For example, it follows from Example 8a that the ith ball withdrawn will be special

with probability k/n, which is intuitively clear, since each of the n balls is equally

likely to be the ith one selected.

EXAMPLE 8b

In Example 8a, let Y

denote the selection number of the ﬁrst special ball withdrawn,

let Y

denote the additional number of balls that are then withdrawn until the second

special ball appears, and, in general, let Y

denote the additional number of balls

withdrawn after the (i − 1)st special ball is selected until the ith is selected, i =

1, ..., k. For instance, if n = 4, k = 2andX

= 1, X

= 0, X

= 1, then

= 1, Y

= 3. Now, Y

= i

, Y

= i

, ..., Y

= i

3 X

= X

= ··· =

+···+i

= 1, X

= 0, otherwise; thus, from the joint mass function of the X

,we

obtain

P{Y

= i

, Y

= i

, ..., Y

= i

k!(n − k)!

+ ··· + i

… n

Hence, the random variables Y

, ..., Y

are exchangeable. Note that it follows from

this result that the number of cards one must select from a well-shufﬂed deck until

284 Chapter 6 Jointly Distributed Random Variables

an ace appears has the same distribution as the number of additional cards one must

select after the ﬁrst ace appears until the next one does, and so on. .

EXAMPLE 8c

The following is known as Polya’s urn model: Suppose that an urn initially contains n

red and m blue balls. At each stage, a ball is randomly chosen, its color is noted, and

it is then replaced along with another ball of the same color. Let X

= 1iftheith ball

selected is red and let it equal 0 if the ith ball is blue, i Ú 1. To obtain a feeling for the

joint probabilities of these X

, note the following special cases:

P{X

= 1, X

= 0, X

= 1, X

= 0}

n + m

n + 1

n + m + 1

n + m + 2

n + 2

n + m + 3

m + 1

n + m + 4

n(n + 1)(n + 2)m(m + 1)

(n + m)(n + m + 1)(n + m + 2)(n + m + 3)(n + m + 4)

and

P{X

= 0, X

= 1, X

= 0, X

= 1, X

= 1}

n + m

n + m + 1

m + 1

n + m + 2

n + 1

n + m + 3

n + 2

n + m + 4

n(n + 1)(n + 2)m(m + 1)

(n + m)(n + m + 1)(n + m + 2)(n + m + 3)(n + m + 4)

By the same reasoning, for any sequence x

, ..., x

that contains r ones and k − r

zeros, we have

P{X

= x

, ..., X

= x

}

n(n + 1) ···(n + r − 1)m(m + 1) ···(m + k − r − 1)

(n + m) ···(n + m + k − 1)

Therefore, for any value of k, the random variables X

, ..., X

are exchangeable.

An interesting corollary of the exchangeability in this model is that the probability

that the ith ball selected is red is the same as the probability that the ﬁrst ball selected

is red, namely,

n+m

. (For an intuitive argument for this initially nonintuitive result,

imagine that all the n + m balls initially in the urn are of different types. That is, one

is a red ball of type 1, one is a red ball of type 2, ..., one is a red ball type of n, one is

a blue ball of type 1, and so on, down to the blue ball of type m. Suppose that when

a ball is selected it is replaced along with another of its type. Then, by symmetry, the

ith ball selected is equally likely to be of any of the n + m distinct types. Because n

of these n + m types are red, the probability is

n+m

.) .

Our ﬁnal example deals with continuous random variables that are exchangeable.

EXAMPLE 8d

Let X

, X

, ..., X

be independent uniform (0, 1) random variables, and denote their

order statistics by X

(1)

, ..., X

(n)

. That is, X

(j)

is the jth smallest of X

, X

, ..., X

Summary 285

Also, let

= X

(1)

= X

(i)

− X

(i−1)

, i = 2, ...n

Show that Y

, ..., Y

are exchangeable.

Solution. The transformations

= x

, ..., y

= x

− x

i−1

i = 2, ..., n

yield

= y

+ ··· + y

i = 1, ..., n

As it is easy to see that the Jacobian of the preceding transformations is equal to 1,

so, from Equation (7.3), we obtain

,...,Y

, y

, ..., y

) = f(y

, y

+ y

, ..., y

+ ··· + y

)

where f is the joint density function of the order statistics. Hence, from Equation (6.1),

we obtain that

,...,Y

, y

, ..., y

) = n!0< y

< y

+ y

< ··· < y

+ ··· + y

< 1

or, equivalently,

,...,Y

, y

, ..., y

) = n!0< y

< 1, i = 1, ..., n, y

+ ··· + y

< 1

Because the preceding joint density is a symmetric function of y

, ..., y

, we see that

the random variables Y

, ..., Y

are exchangeable. .

SUMMARY

The joint cumulative probability distribution function of the pair of random variables

X and Y is deﬁned by

F(x, y) = P{X … x, Y … y}−q < x, y < q

All probabilities regarding the pair can be obtained from F. To ﬁnd the individual

probability distribution functions of X and Y,use

(x) = lim

y→q

F(x, y) F

(y) = lim

x→q

F(x, y)

If X and Y are both discrete random variables, then their joint probability mass

function is deﬁned by

p(i, j) = P{X = i, Y = j}

The individual mass functions are

P{X = i}=



p(i, j) P{Y = j}=



p(i, j)

The random variables X and Y are said to be jointly continuous if there is a func-

tion f (x, y), called the joint probability density function, such that for any two-dimensional

set C,

P{(X, Y) ∈ C}=

f (x, y) dx dy

286 Chapter 6 Jointly Distributed Random Variables

It follows from the preceding formula that

P{x < X < x + dx, y < Y < y + dy} L f (x, y) dx dy

If X and Y are jointly continuous, then they are individually continuous with density

functions

(x) =

−q

f (x, y)dy f

(y) =

−q

f (x, y) dx

The random variables X and Y are independent if, for all sets A and B,

P{X ∈ A, Y ∈ B}=P{X ∈ A}P{Y ∈ B}

If the joint distribution function (or the joint probability mass function in the discrete

case, or the joint density function in the continuous case) factors into a part depending

only on x and a part depending only on y,thenX and Y are independent.

In general, the random variables X

, ..., X

are independent if, for all sets of real

numbers A

, ..., A

P{X

∈ A

, ..., X

∈ A

}=P{X

∈ A

}···P{X

∈ A

}

If X and Y are independent continuous random variables, then the distribution func-

tion of their sum can be obtained from the identity

X+Y

(a) =

−q

(a − y)f

(y)dy

If X

, i = 1, ..., n, are independent normal random variables with respective param-

eters μ

and σ

, i = 1, ..., n,then



i=1

is normal with parameters



i=1

and



i=1

If X

, i = 1, ..., n, are independent Poisson random variables with respective param-

eters λ

, i = 1, ..., n,then



i=1

is Poisson with parameter



i=1

If X and Y are discrete random variables, then the conditional probability mass

function of X given that Y = y is deﬁned by

P{X = x|Y = y}=

p(x, y)

(y)

where p is their joint probability mass function. Also, if X and Y are jointly continu-

ous with joint density function f , then the conditional probability density function of

X given that Y = y is given by

X|Y

(x|y) =

f (x, y)

(y)

The ordered values X

(1)

… X

(2)

… ··· … X

(n)

of a set of independent and identically

distributed random variables are called the order statistics of that set. If the random

variables are continuous and have density function f , then the joint density function

of the order statistics is

f (x

, ..., x

) = n!f(x

) ···f (x

) x

… x

… ··· … x

The random variables X

, ..., X

are called exchangeable if the joint distribution of

, ..., X

is the same for every permutation i

, ..., i

of 1, ..., n.

Problems 287

PROBLEMS

6.1. Two fair dice are rolled. Find the joint probability

mass function of X and Y when

(a) X is the largest value obtained on any die and

Y is the sum of the values;

(b) X is the value on the ﬁrst die and Y is the

larger of the two values;

obtained on the dice.

6.2. Suppose that 3 balls are chosen without replace-

ment from an urn consisting of 5 white and 8 red

balls. Let X

equal 1 if the ith ball selected is white,

and let it equal 0 otherwise. Give the joint proba-

bility mass function of

(a) X

, X

;

(b) X

, X

6.3. In Problem 2, suppose that the white balls are

numbered, and let Y

equal 1 if the ith white ball is

selected and 0 otherwise. Find the joint probability

mass function of

(a) Y

, Y

;

(b) Y

, Y

6.4. Repeat Problem 2 when the ball selected is

replaced in the urn before the next selection.

6.5. Repeat Problem 3a when the ball selected is

replaced in the urn before the next selection.

6.6. A bin of 5 transistors is known to contain 2 that

are defective. The transistors are to be tested, one

at a time, until the defective ones are identiﬁed.

Denote by N

the number of tests made until the

ﬁrst defective is identiﬁed and by N

the number of

additional tests until the second defective is identi-

ﬁed. Find the joint probability mass function of N

and N

6.7. Consider a sequence of independent Bernoulli tri-

als, each of which is a success with probability p.

Let X

be the number of failures preceding the

ﬁrst success, and let X

be the number of failures

between the ﬁrst two successes. Find the joint mass

function of X

and X

6.8. The joint probability density function of X and Y

is given by

f (x, y) = c(y

− x

−y

− y … x … y,0< y < q

(a) Find c.

(b) Find the marginal densities of X and Y.

6.9. The joint probability density function of X and Y

is given by

f (x, y) =





0 < x < 1, 0 < y < 2

(a) Verify that this is indeed a joint density func-

tion.

(b) Compute the density function of X.

(d) Find P{Y >

|X <

(e) Find E[X].

(f) Find E[Y].

6.10. The joint probability density function of X and Y

is given by

f (x, y) = e

−(x+y)

0 … x < q,0 … y < q

Find (a) P{X < Y} and (b) P{X < a}.

6.11. A television store owner ﬁgures that 45 percent of

the customers entering his store will purchase an

ordinary television set, 15 percent will purchase

a plasma television set, and 40 percent will just

be browsing. If 5 customers enter his store on

a given day, what is the probability that he will

sell exactly 2 ordinary sets and 1 plasma set on

that day?

6.12. The number of people that enter a drugstore in

a given hour is a Poisson random variable with

parameter λ = 10. Compute the conditional prob-

ability that at most 3 men entered the drugstore,

given that 10 women entered in that hour. What

assumptions have you made?

6.13. A man and a woman agree to meet at a certain

location about 12:30

P.M. If the man arrives at

a time uniformly distributed between 12:15 and

12:45, and if the woman independently arrives at

a time uniformly distributed between 12:00 and 1

P.M., ﬁnd the probability that the ﬁrst to arrive

waits no longer than 5 minutes. What is the proba-

bility that the man arrives ﬁrst?

6.14. An ambulance travels back and forth at a con-

stant speed along a road of length L.Atacertain

moment of time, an accident occurs at a point uni-

formly distributed on the road. [That is, the dis-

tance of the point from one of the ﬁxed ends of the

road is uniformly distributed over (0, L).] Assum-

ing that the ambulance’s location at the moment

of the accident is also uniformly distributed, and

assuming independence of the variables, compute

the distribution of the distance of the ambulance

from the accident.

6.15. The random vector (X, Y)issaidtobeuniformly

distributed over a region R in the plane if, for some

constant c, its joint density is

f (x, y) =

c if(x, y) ∈ R

0 otherwise

288 Chapter 6 Jointly Distributed Random Variables

(a) Show that 1/c = area of region R.

Suppose that (X, Y) is uniformly distributed over

the square centered at (0, 0) and with sides of

length 2.

(b) Show that X and Y are independent, with

each being distributed uniformly over (−1, 1).

circle of radius 1 centered at the origin? That

is, ﬁnd P{X

+ Y

… 1}.

6.16. Suppose that n points are independently chosen at

random on the circumference of a circle, and we

want the probability that they all lie in some semi-

circle. That is, we want the probability that there is

a line passing through the center of the circle such

that all the points are on one side of that line, as

shown in the following diagram:

Let P

, ..., P

denote the n points. Let A denote

the event that all the points are contained in some

semicircle, and let A

be the event that all the

points lie in the semicircle beginning at the point

and going clockwise for 180

◦

, i = 1, ..., n.

(a) Express A in terms of the A

(b) Are the A

mutually exclusive?

6.17. Three points X

, X

are selected at random

on a line L. What is the probability that X

lies

between X

and X

6.18. Two points are selected randomly on a line of

length L so as to be on opposite sides of the mid-

point of the line. [In other words, the two points

X and Y are independent random variables such

that X is uniformly distributed over (0, L/2) and

Y is uniformly distributed over (L/2, L).] Find

the probability that the distance between the two

points is greater than L/3.

6.19. Show that f (x, y) = 1/x,0 < y < x < 1, is a joint

density function. Assuming that f is the joint den-

sity function of X, Y, ﬁnd

(a) the marginal density of Y;

(b) the marginal density of X;

6.20. The joint density of X and Y is given by

f (x, y) =

−(x+y)

x > 0, y > 0

0 otherwise

Are X and Y independent? If, instead, f (x, y)were

given by

f (x, y) =

20< x < y,0< y < 1

0 otherwise

would X and Y be independent?

6.21. Let

f (x, y) = 24xy 0 … x … 1, 0 … y … 1, 0 … x + y … 1

and let it equal 0 otherwise.

(a) Show that f (x, y) is a joint probability density

function.

(b) Find E[X].

6.22. The joint density function of X and Y is

f (x,

y) =

x + y 0 < x < 1, 0 < y < 1

0 otherwise

(a) Are X and Y independent?

(b) Find the density function of X.

6.23. The random variables X and Y have joint density

function

f (x, y) = 12xy(1 − x) 0 < x < 1, 0 < y < 1

and equal to 0 otherwise.

(a) Are X and Y independent?

(b) Find E[X].

(d) Find Var(X).

(e) Find Var(Y).

6.24. Consider independent trials, each of which results

in outcome i, i = 0, 1, ..., k, with probability



i=0

= 1. Let N denote the number of trials

needed to obtain an outcome that is not equal to 0,

and let X be that outcome.

(a) Find P{N = n}, n Ú 1.

(b) Find P{X = j}, j = 1, ..., k.

n}P{X = j}.

(d) Is it intuitive to you that N is independent

of X?

(e) Is it intuitive to you that X is independent

of N?

6.25. Suppose that 10

people arrive at a service station

at times that are independent random variables,

Problems 289

each of which is uniformly distributed over (0, 10

Let N denote the number that arrive in the ﬁrst

hour. Find an approximation for P{N = i}.

6.26. Suppose that A, B, C, are independent random

variables, each being uniformly distributed over

(0, 1).

(a) What is the joint cumulative distribution func-

tion of A, B, C?

(b) What is the probability that all of the roots of

the equation Ax

+ Bx + C = 0 are real?

6.27. If X

and X

are independent exponential ran-

dom variables with respective parameters λ

and

, ﬁnd the distribution of Z = X

. Also com-

pute P{X

< X

6.28. The time that it takes to service a car is an expo-

nential random variable with rate 1.

(a) If A. J. brings his car in at time 0 and M. J.

brings her car in at time t, what is the probabil-

ity that M. J.’s car is ready before A. J.’s car?

(Assume that service times are independent

and service begins upon arrival of the car.)

(b) If both cars are brought in at time 0, with

work starting on M. J.’s car only when A. J.’s

car has been completely serviced, what is the

probability that M. J.’s car is ready before

time 2?

6.29. The gross weekly sales at a certain restaurant is

a normal random variable with mean $2200 and

standard deviation $230. What is the probabil-

ity that

(a) the total gross sales over the next 2 weeks

exceeds $5000;

(b) weekly sales exceed $2000 in at least 2 of the

next 3 weeks?

What independence assumptions have you made?

6.30. Jill’s bowling scores are approximately normally

distributed with mean 170 and standard deviation

20, while Jack’s scores are approximately normally

distributed with mean 160 and standard deviation

15. If Jack and Jill each bowl one game, then

assuming that their scores are independent ran-

dom variables, approximate the probability that

(a) Jack’s score is higher;

(b) the total of their scores is above 350.

6.31. According to the U.S. National Center for Health

Statistics, 25.2 percent of males and 23.6 percent of

females never eat breakfast. Suppose that random

samples of 200 men and 200 women are chosen.

Approximate the probability that

(a) at least 110 of these 400 people never eat

breakfast;

(b) the number of the women who never eat

breakfast is at least as large as the number of

the men who never eat breakfast.

6.32. The expected number of typographical errors on a

page of a certain magazine is .2. What is the prob-

ability that an article of 10 pages contains (a) 0 and

(b) 2 or more typographical errors? Explain your

reasoning!

6.33. The monthly worldwide average number of air-

plane crashes of commercial airlines is 2.2. What

is the probability that there will be

(a) more than 2 such accidents in the next month?

(b) more than 4 such accidents in the next 2

months?

Explain your reasoning!

6.34. Jay has two jobs to do, one after the other. Each

attempt at job i takes one hour and is successful

with probability p

. If p

= .3andp

= .4, what

is the probability that it will take Jay more than 12

hours to be successful on both jobs?

6.35. In Problem 4, calculate the conditional probability

mass function of X

given that

(a) X

= 1;

(b) X

= 0.

6.36. In Problem 3, calculate the conditional probability

mass function of Y

given that

(a) Y

= 1;

(b) Y

= 0.

6.37. In Problem 5, calculate the conditional probability

mass function of Y

given that

(a) Y

= 1;

(b) Y

= 0.

6.38. Choose a number X at random from the set of

numbers {1, 2, 3, 4, 5}. Now choose a number at

random from the subset no larger than X,thatis,

from {1, ..., X}. Call this second number Y.

(a) Find the joint mass function of X and Y.

(b) Find the conditional mass function of X given

that Y = i.Doitfori = 1, 2, 3, 4, 5.

6.39. Two dice are rolled. Let X and Y denote, respec-

tively, the largest and smallest values obtained.

Compute the conditional mass function of Y given

X = i,fori = 1, 2, ...,6. Are X and Y indepen-

dent? Why?

6.40. The joint probability mass function of X and Y is

given by

p(1, 1) =

p(1, 2) =

p(2, 1) =

p(2, 2) =

(a) Compute the conditional mass function of X

given Y = i, i = 1, 2.

(b) Are X and Y independent?

290 Chapter 6 Jointly Distributed Random Variables

P{X/Y > 1}.

6.41. The joint density function of X and Y is given by

f (x, y) = xe

−x(y+1)

x > 0, y > 0

(a) Find the conditional density of X, given Y =y,

and that of Y, given X = x.

(b) Find the density function of Z = XY.

6.42. The joint density of X and Y is

f (x, y) = c(x

− y

−x

0 … x < q, −x … y … x

Find the conditional distribution of Y, given

X = x.

6.43. An insurance company supposes that each per-

son has an accident parameter and that the yearly

number of accidents of someone whose accident

parameter is λ is Poisson distributed with mean λ.

They also suppose that the parameter value of a

newly insured person can be assumed to be the

value of a gamma random variable with param-

eters s and α. If a newly insured person has n

accidents in her ﬁrst year, ﬁnd the conditional den-

sity of her accident parameter. Also, determine the

expected number of accidents that she will have in

the following year.

6.44. If X

, X

are independent random variables

that are uniformly distributed over (0, 1), com-

pute the probability that the largest of the three

is greater than the sum of the other two.

6.45. A complex machine is able to operate effectively

as long as at least 3 of its 5 motors are functioning.

If each motor independently functions for a ran-

dom amount of time with density function f (x) =

−x

, x > 0, compute the density function of the

length of time that the machine functions.

6.46. If 3 trucks break down at points randomly dis-

tributed on a road of length L, ﬁnd the probability

that no 2 of the trucks are within a distance d of

each other when d … L/2.

6.47. Consider a sample of size 5 from a uniform distri-

bution over (0, 1). Compute the probability that

the median is in the interval





6.48. If X

, X

are independent and iden-

tically distributed exponential random variables

with the parameter λ, compute

(a) P{min(X

, ..., X

) … a};

(b) P{max(X

, ..., X

) … a}.

6.49. Let X

(1)

, X

(2)

, ..., X

(n)

be the order statistics

of a set of n independent uniform (0, 1) random

variables. Find the conditional distribution of X

(n)

given that X

(1)

= s

, X

(2)

= s

, ..., X

(n−1)

n−1

6.50. Let Z

and Z

be independent standard normal

random variables. Show that X, Y has a bivariate

normal distribution when X = Z

, Y = Z

+ Z

6.51. Derive the distribution of the range of a sample of

size 2 from a distribution having density function

f (x) = 2x,0< x < 1.

6.52. Let X and Y denote the coordinates of a point uni-

formly chosen in the circle of radius 1 centered at

the origin. That is, their joint density is

f (x, y) =

+ y

… 1

Find the joint density function of the polar coordi-

nates R = (X

+ Y

)

1/2

and  = tan

−1

Y/X.

6.53. If X and Y are independent random variables

both uniformly distributed over (0, 1), ﬁnd the

joint density function of R =

+ Y

, Θ =

tan

−1

Y/X.

6.54. If U is uniform on (0, 2π) and Z, independent of

U, is exponential with rate 1, show directly (with-

out using the results of Example 7b) that X and Y

deﬁned by

X =

√

2Z cos U

Y =

√

2Z sin U

are independent standard normal random vari-

ables.

6.55. X and Y have joint density function

f (x, y) =

x Ú 1, y Ú 1

(a) Compute the joint density function of U =

XY, V = X/Y.

(b) What are the marginal densities?

6.56. If X and Y are independent and identically dis-

tributed uniform random variables on (0, 1),

compute the joint density of

(a) U = X + Y, V = X/Y;

(b) U = X, V = X/Y;

6.57. Repeat Problem 6.56 when X and Y are inde-

pendent exponential random variables, each with

parameter λ = 1.

6.58. If X

and X

are independent exponential random

variables, each having parameter λ, ﬁnd the joint

density function of Y

= X

+ X

and Y

= e

6.59. If X, Y,andZ are independent random variables

having identical density functions f (x) = e

−x

,0 <

x < q, derive the joint distribution of U = X +

Y, V = X + Z, W = Y + Z.

Theoretical Exercises 291

6.60. In Example 8b, let Y

k+1

= n + 1 −



i=1

. Show

that Y

, ..., Y

, Y

k+1

are exchangeable. Note that

k+1

is the number of balls one must observe to

obtain a special ball if one considers the balls in

their reverse order of withdrawal.

6.61. Consider an urn containing n balls numbered

1, ..., n, and suppose that k of them are randomly

withdrawn. Let X

equal 1 if ball number i is

removed and let X

be 0 otherwise. Show that

, ..., X

are exchangeable.

THEORETICAL EXERCISES

6.1. Verify Equation (1.2).

6.2. Suppose that the number of events occurring in

a given time period is a Poisson random vari-

able with parameter λ. If each event is classi-

ﬁed as a type i event with probability p

, i =

1, ..., n,



= 1, independently of other events,

show that the numbers of type i events that occur,

i = 1, ..., n, are independent Poisson random vari-

ables with respective parameters λp

, i = 1, ..., n.

6.3. Suggest a procedure for using Buffon’s needle

problem to estimate π . Surprisingly enough, this

was once a common method of evaluating π.

6.4. Solve Buffon’s needle problem when L > D.

ANSWER:

πD

(1 − sin θ) + 2θ/π,wherecosθ =

D/L.

6.5. If X and Y are independent continuous positive

random variables, express the density function of

(a) Z = X/Y and (b) Z = XY in terms of the

density functions of X and Y. Evaluate the density

functions in the special case where X and Y are

both exponential random variables.

6.6. If X and Y are jointly continuous with joint density

function f

X,Y

(x, y), show that X + Y is continuous

with density function

X+Y

(t) =

−q

X,Y

(x, t − x) dx

6.7. (a) If X has a gamma distribution with parame-

ters (t, λ), what is the distribution of cX, c > 0?

(b) Show that

2λ

has a gamma distribution with parameters n, λ

when n is a positive integer and χ

is a

chi-squared random variable with 2n degrees

of freedom.

6.8. Let X and Y be independent continuous ran-

dom variables with respective hazard rate func-

tions λ

(t) and λ

(t),andsetW = min(X, Y).

(a) Determine the distribution function of W in

terms of those of X and Y.

(b) Show that λ

(t), the hazard rate function of

W, is given by

(t) = λ

(t) + λ

(t)

6.9. Let X

, ..., X

be independent exponential ran-

dom variables having a common parameter λ.

Determine the distribution of min(X

, ..., X

6.10. The lifetimes of batteries are independent expo-

nential random variables, each having parame-

ter λ. A ﬂashlight needs 2 batteries to work. If one

has a ﬂashlight and a stockpile of n batteries, what

is the distribution of time that the ﬂashlight can

operate?

6.11. Let X

, X

be independent continu-

ous random variables having a common distribu-

tion function F and density function f,andset

I = P{X

< X

}

(a) Show that I does not depend on F.

Hint:WriteI as a ﬁve-dimensional integral and

make the change of variables u

= F(x

), i =

1, ...,5.

(b) Evaluate I.

to (b).

6.12. Show that the jointly continuous (discrete) ran-

dom variables X

, ..., X

are independent if and

only if their joint probability density (mass) func-

tion f (x

, ..., x

) can be written as

f (x

, ..., x

) =



i=1

)

for nonnegative functions g

(x), i = 1, ..., n.

6.13. In Example 5c we computed the conditional den-

sity of a success probability for a sequence of tri-

als when the ﬁrst n + m trials resulted in n suc-

cesses. Would the conditional density change if

we speciﬁed which n of these trials resulted in

successes?

292 Chapter 6 Jointly Distributed Random Variables

6.14. Suppose that X and Y are independent geometric

random variables with the same parameter p.

(a) Without any computations, what do you think

is the value of

P{X = i|X + Y = n}?

Hint: Imagine that you continually ﬂip a coin hav-

ing probability p of coming up heads. If the second

head occurs on the nth ﬂip, what is the probability

mass function of the time of the ﬁrst head?

(b) Verify your conjecture in part (a).

6.15. Consider a sequence of independent trials, with

each trial being a success with probability p. Given

that the kth success occurs on trial n, show that

all possible outcomes of the ﬁrst n − 1trialsthat

consist of k − 1 successes and n − k failures are

equally likely.

6.16. If X and Y are independent binomial random

variables with identical parameters n and p, show

analytically that the conditional distribution of X

given that X + Y = m is the hypergeometric dis-

tribution. Also, give a second argument that yields

the same result without any computations.

Hint: Suppose that 2n coins are ﬂipped. Let X

denote the number of heads in the ﬁrst n ﬂips and

Y the number in the second n ﬂips. Argue that

given a total of m heads, the number of heads in

the ﬁrst n ﬂips has the same distribution as the

number of white balls selected when a sample of

size m is chosen from n white and n black balls.

6.17. Suppose that X

, i = 1, 2, 3 are independent Pois-

son random variables with respective means λ

, i =

1, 2, 3. Let X = X

+ X

and Y = X

+ X

The random vector X, Y is said to have a bivari-

ate Poisson distribution. Find its joint probability

mass function. That is, ﬁnd P{X = n, Y = m}.

6.18. Suppose X and Y are both integer-valued random

variables. Let

p(i|j) = P(X = i|Y = j)

and

q(j|i) = P(Y = j|X = i)

Show that

P(X = i, Y = j) =

p(i|j)



p(i|j)

q(j|i)

6.19. Let X

, X

be independent and identically dis-

tributed continuous random variables. Compute

(a) P{X

> X

};

(b) P{X

> X

< X

};

> X

};

(d) P{X

> X

< X

6.20. Let U denote a random variable uniformly dis-

tributed over (0, 1). Compute the conditional dis-

tribution of U given that

(a) U > a;

(b) U < a;

where 0 < a < 1.

6.21. Suppose that W, the amount of moisture in the

air on a given day, is a gamma random vari-

able with parameters (t, β). That is, its density is

f (w) = βe

−βw

(βw)

t−1

/(t), w > 0. Suppose also

that given that W = w, the number of accidents

during that day—call it N—has a Poisson distribu-

tion with mean w. Show that the conditional dis-

tribution of W given that N = n is the gamma

distribution with parameters (t + n, β + 1).

6.22. Let W be a gamma random variable with param-

eters (t, β), and suppose that conditional on

W = w, X

, X

, ..., X

are independent exponen-

tial random variables with rate w. Show that the

conditional distribution of W given that X

, X

= x

, ..., X

= x

is gamma with parame-

ters



t + n, β +



i=1



6.23. A rectangular array of mn numbers arranged in n

rows, each consisting of m columns, is said to con-

tain a saddlepoint if there is a number that is both

the minimum of its row and the maximum of its

column. For instance, in the array

132

0 −26

.5123

the number 1 in the ﬁrst row, ﬁrst column is a

saddlepoint. The existence of a saddlepoint is of

signiﬁcance in the theory of games. Consider a

rectangular array of numbers as described previ-

ously and suppose that there are two individuals—

A and B—that are playing the following game: A

is to choose one of the numbers 1, 2, ..., n and B

one of the numbers 1, 2, ..., m. These choices are

announced simultaneously, and if A chose i and

B chose j,thenA wins from B the amount spec-

iﬁed by the number in the ith row, jth column of

the array. Now suppose that the array contains

a saddlepoint—say the number in the row r and

column k—call this number x

. Now if player A

chooses row r, then that player can guarantee her-

self a win of at least x

(since x

is the minimum

number in the row r). On the other hand, if player

B chooses column k, then he can guarantee that he

will lose no more than x

(since x

is the maxi-

mum number in the column k). Hence, as A has a

way of playing that guarantees her a win of x

and

as B has a way of playing that guarantees he will

lose no more than x

, it seems reasonable to take

Self-Test Problems and Exercises 293

these two strategies as being optimal and declare

that the value of the game to player A is x

If the nm numbers in the rectangular array

described are independently chosen from an

arbitrary continuous distribution, what is the

probability that the resulting array will contain a

saddle-point?

6.24. If X is exponential with rate λ, ﬁnd P {[X] = n, X −

[X] … x},where[x] is deﬁned as the largest integer

less than or equal to x. Can you conclude that [X]

and X − [X] are independent?

6.25. Suppose that F(x) is a cumulative distribution

function. Show that (a) F

(x) and (b) 1 − [1 −

F(x)]

are also cumulative distribution functions

when n is a positive integer.

Hint:LetX

, ..., X

be independent random vari-

ables having the common distribution function F.

Deﬁne random variables Y and Z in terms of the

so that P{Y … x}=F

(x) and P{Z … x}=

1 − [1 − F(x)]

6.26. Show that if n people are distributed at random

along a road L miles long, then the probability that

no 2 people are less than a distance D miles apart

is when D … L/(n − 1),[1 − (n − 1)D/L]

. What

if D > L/(n − 1)?

6.27. Establish Equation (6.2) by differentiating Equa-

tion (6.4).

6.28. Show that the median of a sample of size 2n + 1

from a uniform distribution on (0, 1) has a beta

distribution with parameters (n + 1, n + 1).

6.29. Verify Equation (6.6), which gives the joint density

of X

(i)

and X

(j)

6.30. Compute the density of the range of a sample of

size n from a continuous distribution having den-

sity function f .

6.31. Let X

(1)

… X

(2)

… ···… X

(n)

be the ordered values

of n independent uniform (0, 1) random variables.

Prove that for 1 … k … n + 1,

P{X

(k)

− X

(k−1)

> t}=(1 − t)

where X

(0)

K 0, X

(n+1)

K t.

6.32. Let X

, ..., X

be a set of independent and identi-

cally distributed continuous random variables hav-

ing distribution function F,andletX

(i)

, i = 1, ..., n

denote their ordered values. If X, independent of

the X

, i = 1, ..., n, also has distribution F, deter-

mine

(a) P{X > X

(n)

};

(b) P{X > X

(1)

};

(i)

< X < X

(j)

},1 … i < j … n.

6.33. Let X

, ..., X

be independent and identically

distributed random variables having distribution

function F and density f . The quantity M K [X

(1)

(n)

]/2, deﬁned to be the average of the small-

est and largest values in X

, ..., X

, is called the

midrange of the sequence. Show that its distribu-

tion function is

(m) = n

−q

[F(2m − x) − F(x)]

n−1

f (x) dx

6.34. Let X

, ..., X

be independent uniform (0, 1) ran-

dom variables. Let R = X

(n)

− X

(1)

denote the

range and M = [X

(n)

+ X

(1)

]/2 the midrange of

, ..., X

. Compute the joint density function of

R and M.

6.35. If X and Y are independent standard normal ran-

dom variables, determine the joint density func-

tion of

U = XV=

Then use your result to show that X/Y has a

Cauchy distribution.

SELF-TEST PROBLEMS AND EXERCISES

6.1. Each throw of an unfair die lands on each of the

odd numbers 1, 3, 5 with probability C and on each

of the even numbers with probability 2C.

(a) Find C.

(b) Suppose that the die is tossed. Let X equal

1 if the result is an even number, and let it

be 0 otherwise. Also, let Y equal 1 if the

result is a number greater than three and

let it be 0 otherwise. Find the joint prob-

ability mass function of X and Y. Suppose

now that 12 independent tosses of the die are

made.

comes occurs exactly twice.

(d) Find the probability that 4 of the outcomes

are either one or two, 4 are either three or

four, and 4 are either ﬁve or six.

(e) Find the probability that at least 8 of the

tosses land on even numbers.

6.2. The joint probability mass function of the random

variables X, Y, Z is

p(1, 2, 3) = p(2, 1, 1) = p(2, 2, 1) = p(2, 3, 2) =

Find (a) E[XYZ], and (b) E[XY + XZ + YZ].

6.3. The joint density of X and Y is given by

f (x, y) = C(y − x)e

−y

− y < x < y,0< y < q

294 Chapter 6 Jointly Distributed Random Variables

(a) Find C.

(b) Find the density function of X.

(d) Find E[X].

(e) Find E[Y].

6.4. Let r = r

+ ... + r

,whereallr

are positive

integers. Argue that if X

, ..., X

has a multino-

mial distribution, then so does Y

, ..., Y

where,

with r

= 0,

i−1



j=r

i−1

, i … k

That is, Y

is the sum of the ﬁrst r

of the X



s, Y

is the sum of the next r

, and so on.

6.5. Suppose that X, Y,andZ are independent random

variables that are each equally likely to be either

1 or 2. Find the probability mass function of (a)

XYZ,(b)XY + XZ + YZ,and(c)X

+ YZ.

6.6. Let X and Y be continuous random variables with

joint density function

f (x, y) =

⎧

⎨

⎩

+ cy 0 < x < 1, 1 < y < 5

0 otherwise

where c is a constant.

(a) What is the value of c?

(b) Are X and Y independent?

6.7. The joint density function of X and Y is

f (x, y) =

xy 0 < x < 1, 0 < y < 2

0 otherwise

(a) Are X and Y independent?

(b) Find the density function of X.

(d) Find the joint distribution function.

(e) Find E[Y].

(f) Find P{X + Y < 1}.

6.8. Consider two components and three types of

shocks. A type 1 shock causes component 1 to fail,

a type 2 shock causes component 2 to fail, and a

type 3 shock causes both components 1 and 2 to

fail. The times until shocks 1, 2, and 3 occur are

independent exponential random variables with

respective rates λ

, λ

,andλ

.LetX

denote the

time at which component i fails, i = 1, 2. The

random variables X

, X

are said to have a joint

bivariate exponential distribution. Find P{X

s, X

> t}.

6.9. Consider a directory of classiﬁed advertisements

that consists of m pages, where m is very large.

Suppose that the number of advertisements per

page varies and that your only method of ﬁnding

out how many advertisements there are on a spec-

iﬁed page is to count them. In addition, suppose

that there are too many pages for it to be feasi-

ble to make a complete count of the total num-

ber of advertisements and that your objective is

to choose a directory advertisement in such a way

that each of them has an equal chance of being

selected.

(a) If you randomly choose a page and then

randomly choose an advertisement from that

page, would that satisfy your objective? Why

or why not?

Let n(i) denote the number of advertise-

ments on page i, i = 1, ..., m, and suppose

that whereas these quantities are unknown,

we can assume that they are all less than

or equal to some speciﬁed value n. Con-

sider the following algorithm for choosing an

advertisement.

Step 1. Choose a page at random. Suppose it is

page X. Determine n(X) by counting the

number of advertisements on page X.

Step 2. “Accept” page X with probability n(X)/n.

If page X is accepted, go to step 3. Other-

wise, return to step 1.

Step 3. Randomly choose one of the advertise-

ments on page X.

Call each pass of the algorithm through step

1 an iteration. For instance, if the ﬁrst ran-

domly chosen page is rejected and the second

accepted, than we would have needed 2 itera-

tions of the algorithm to obtain an advertise-

ment.

(b) What is the probability that a single iteration

of the algorithm results in the acceptance of

an advertisement on page i?

of the algorithm results in the acceptance of

an advertisement?

(d) What is the probability that the algorithm

goes through k iterations, accepting the jth

advertisement on page i on the ﬁnal iteration?

(e)

What is the probability that the jth advertise-

ment on page i is the advertisement obtained

from the algorithm?

(f) What is the expected number of iterations

taken by the algorithm?

6.10. The “random” parts of the algorithm in Self-Test

Problem 8 can be written in terms of the generated

values of a sequence of independent uniform (0,

1) random variables, known as random numbers.

With [x] deﬁned as the largest integer less than or

equal to x, the ﬁrst step can be written as follows:

Self-Test Problems and Exercises 295

Step 1. Generate a uniform (0, 1) random variable U.

Let X = [mU] + 1, and determine the value of

n(X).

(a) Explain why the above is equivalent to step 1

of Problem 8.

Hint: What is the probability mass function of X?

(b) Write the remaining steps of the algorithm in

a similar style.

6.11. Let X

, X

, ... be a sequence of independent uni-

form (0, 1) random variables. For a ﬁxed con-

stant c, deﬁne the random variable N by

N = min{n : X

> c}

Is N independent of X

? That is, does know-

ing the value of the ﬁrst random variable that is

greater than c affect the probability distribution of

when this random variable occurs? Give an intu-

itive explanation for your answer.

6.12. The accompanying dartboard is a square whose

sides are of length 6:

The three circles are all centered at the center of

the board and are of radii 1, 2, and 3, respectively.

Darts landing within the circle of radius 1 score 30

points, those landing outside this circle, but within

the circle of radius 2, are worth 20 points, and

those landing outside the circle of radius 2, but

within the circle of radius 3, are worth 10 points.

Darts that do not land within the circle of radius

3 do not score any points. Assuming that each

dart that you throw will, independently of what

occurred on your previous throws, land on a point

uniformly distributed in the square, ﬁnd the prob-

abilities of the accompanying events:

(a) You score 20 on a throw of the dart.

(b) You score at least 20 on a throw of the

dart.

(d) The expected value of your score on a throw

of the dart.

(e) Both of your ﬁrst two throws score at least 10.

(f) Your total score after two throws is 30.

6.13. A model proposed for NBA basketball supposes

that when two teams with roughly the same record

play each other, the number of points scored in

a quarter by the home team minus the number

scored by the visiting team is approximately a nor-

mal random variable with mean 1.5 and variance

6. In addition, the model supposes that the point

differentials for the four quarters are independent.

Assume that this model is correct.

(a) What is the probability that the home team

wins?

(b) What is the conditional probability that the

home team wins, given that it is behind by 5

points at halftime?

home team wins, given that it is ahead by 5

points at the end of the ﬁrst quarter?

6.14. Let N be a geometric random variable with param-

eter p. Suppose that the conditional distribution

of X given that N = n is the gamma distribu-

tion with parameters n and λ. Find the condi-

tional probability mass function of N given that

X = x.

6.15. Let X and Y be independent uniform (0, 1) ran-

dom variables.

(a) Find the joint density of U = X, V = X + Y.

(b) Use the result obtained in part (a) to compute

the density function of V.

6.16. You and three other people are to place bids for an

object, with the high bid winning. If you win, you

plan to sell the object immediately for 10 thousand

dollars. How much should you bid to maximize

your expected proﬁt if you believe that the bids

of the others can be regarded as being indepen-

dent and uniformly distributed between 7 and 11

thousand dollars?

6.17.

Find the probability that X

, X

, ..., X

is a per-

mutation of 1, 2, ..., n,whenX

, X

, ..., X

are

independent and

(a) each is equally likely to be any of the values

1, ..., n;

(b) each has the probability mass function P{X

j}=p

, j = 1, ..., n.

6.18. Let X

, ..., X

and Y

, ..., Y

be independent ran-

dom vectors, with each vector being a random

ordering of k ones and n − k zeroes. That is, their

joint probability mass functions are

P{X

= i

, ..., X

= i

}=P{Y

= i

, ..., Y

= i

}





, i

= 0, 1,



j=1

= k

Let

N =



i=1

− Y

296 Chapter 6 Jointly Distributed Random Variables

denote the number of coordinates at which the two

vectors have different values. Also, let M denote

the number of values of i for which X

= 1, Y

= 0.

(a) Relate N to M.

(b) What is the distribution of M?

(d) Find Var(N).

∗

6.19. Let Z

, Z

, ..., Z

be independent standard nor-

mal random variables, and let



i=1

(a) What is the conditional distribution of S

given that S

= y,fork = 1, ..., n?

(b) Show that, for 1 … k … n, the conditional

distribution of S

given that S

= x is

normal with mean xk/n and variance k(n −

k)/n.

6.20. Let X

, X

, ... be a sequence of independent and

identically distributed continuous random vari-

ables. Find

(a) P{X

> X

= max(X

, ..., X

)}

(b) P{X

> X

= max(X

, ..., X

)}

CHAPTER 7

Properties of Expectation

7.1 INTRODUCTION

7.2 EXPECTATION OF SUMS OF RANDOM VARIABLES

7.3 MOMENTS OF THE NUMBER OF EVENTS THAT OCCUR

7.4 COVARIANCE, VARIANCE OF SUMS, AND CORRELATIONS

7.5 CONDITIONAL EXPECTATION

7.6 CONDITIONAL EXPECTATION AND PREDICTION

7.7 MOMENT GENERATING FUNCTIONS

7.8 ADDITIONAL PROPERTIES OF NORMAL RANDOM VARIABLES

7.9 GENERAL DEFINITION OF EXPECTATION

7.1 INTRODUCTION

In this chapter, we develop and exploit additional properties of expected values. To

begin, recall that the expected value of the random variable X is deﬁned by

E[X] =



xp(x)

where X is a discrete random variable with probability mass function p(x), and by

E[X] =

−q

xf (x) dx

when X is a continuous random variable with probability density function f(x).

Since E[X] is a weighted average of the possible values of X, it follows that if X

must lie between a and b, then so must its expected value. That is, if

P{a … X … b}=1

then

a … E[X] … b

To verify the preceding statement, suppose that X is a discrete random variable for

which P{a … X … b}=1. Since this implies that p(x) = 0 for all x outside of the

interval [a, b], it follows that

E[X] =



x:p(x)>0

xp(x)



x:p(x)>0

ap(x)

297

298 Chapter 7 Properties of Expectation

= a



x:p(x)>0

p(x)

= a

In the same manner, it can be shown that E[X] … b, so the result follows for discrete

random variables. As the proof in the continuous case is similar, the result follows.

7.2 EXPECTATION OF SUMS OF RANDOM VARIABLES

For a two-dimensional analog of Propositions 4.1 of Chapter 4 and 2.1 of Chapter 5,

which give the computational formulas for the expected value of a function of a ran-

dom variable, suppose that X and Y are random variables and g is a function of two

variables. Then we have the following result.

Proposition 2.1. If X and Y have a joint probability mass function p(x,y),then

E[g(X, Y)] =



g(x, y)p(x, y)

If X and Y have a joint probability density function f(x,y),then

E[g(X, Y)] =

−q

g(x, y)f (x, y) dx dy

Let us give a proof of Proposition 2.1 when the random variables X and Y are

jointly continuous with joint density function f (x, y) and when g(X, Y) is a nonneg-

ative random variable. Because g(X, Y) Ú 0, we have, by Lemma 2.1 of Chapter 5,

that

E[g(X, Y)] =

P{g(X, Y)>t}dt

Writing

P{g(X, Y)>t}=

(x,y):g(x,y)>t

f (x, y) dy dx

shows that

E[g(X, Y)] =

(x,y):g(x,y)>t

f (x, y) dy dx dt

Interchanging the order of integration gives

E[g(X, Y) =

g(x,y)

t=0

f (x, y) dt dy dx

g(x, y) f (x, y) dy dx

Thus, the result is proven when g(X, Y) is a nonnegative random variable. The gen-

eral case then follows as in the one-dimensional case. (See Theoretical Exercises 2

and 3 of Chapter 5.)

EXAMPLE 2a

An accident occurs at a point X that is uniformly distributed on a road of length L.

At the time of the accident, an ambulance is at a location Y that is also uniformly

Section 7.2 Expectation of Sums of Random Variables 299

distributed on the road. Assuming that X and Y are independent, ﬁnd the expected

distance between the ambulance and the point of the accident.

Solution. We need to compute E[|X − Y|]. Since the joint density function of X and

Y is

f (x, y) =

,0< x < L,0< y < L

it follows from Proposition 2.1 that

E[|X − Y|] =

|x − y|dy dx

Now,

|x − y|dy =

(x − y)dy +

(y − x)dy

−

− x(L − x)

+ x

− xL

Therefore,

E[|X − Y|] =



+ x

− xL



For an important application of Proposition 2.1, suppose that E[X]andE[Y] are

both ﬁnite and let g(X, Y) = X + Y. Then, in the continuous case,

E[X + Y] =

−q

(x + y)f (x, y) dx dy

−q

xf (x, y) dy dx +

−q

yf (x, y) dx dy

−q

(x) dx +

−q

(y) dy

= E[X] + E[Y]

The same result holds in general; thus, whenever E[X]andE[Y] are ﬁnite,

E[X + Y] = E[X] + E [Y] (2.1)

EXAMPLE 2b

Suppose that, for random variables X and Y,

X Ú Y

That is, for any outcome of the probability experiment, the value of the random vari-

able X is greater than or equal to the value of the random variable Y. Since x Ú y is

equivalent to the inequality X − Y Ú 0, it follows that E[X − Y] Ú 0, or, equivalently,

E[X] Ú E[Y] .

300 Chapter 7 Properties of Expectation

Using Equation (2.1), we may show by a simple induction proof that if E[X

]is

ﬁnite for all i = 1, ..., n,then

E[X

+ ··· + X

] = E[X

] + ··· + E[X

] (2.2)

Equation (2.2) is an extremely useful formula whose utility will now be illustrated by

a series of examples.

EXAMPLE 2c The sample mean

Let X

, ..., X

be independent and identically distributed random variables having

distribution function F and expected value μ. Such a sequence of random variables is

said to constitute a sample from the distribution F. The quantity

X =



i=1

is called the sample mean. Compute E[

X].

Solution.

X] = E

⎡

⎣



i=1

⎤

⎦

⎡

⎣



i=1

⎤

⎦



i=1

E[X

]

= μ since E[X

] K μ

That is, the expected value of the sample mean is μ, the mean of the distribution.

When the distribution mean μ is unknown, the sample mean is often used in statistics

to estimate it. .

EXAMPLE 2d Boole’s inequality

Let A

, ..., A

denote events, and deﬁne the indicator variables X

, i = 1, ..., n,by

1ifA

occurs

0 otherwise

Let

X =



i=1

so X denotes the number of the events A

that occur. Finally, let

Y =

1ifX Ú 1

0 otherwise

Section 7.2 Expectation of Sums of Random Variables 301

so Y is equal to 1 if at least one of the A

occurs and is 0 otherwise. Now, it is imme-

diate that

X Ú Y

E[X] Ú E[Y]

But since

E[X] =



i=1

E[X

] =



i=1

P(A

)

and

E[Y] = P{at least one of the A

occur}=P

⎛

⎝



i=1

⎞

⎠

we obtain Boole’s inequality, namely,

⎛

⎝



i=1

⎞

⎠

…



i=1

P(A

) .

The next three examples show how Equation (2.2) can be used to calculate the

expected value of binomial, negative binomial, and hypergeometric random vari-

ables. These derivations should be compared with those presented in Chapter 4.

EXAMPLE 2e Expectation of a binomial random variable

Let X be a binomial random variable with parameters n and p. Recalling that such

a random variable represents the number of successes in n independent trials when

each trial has probability p of being a success, we have that

X = X

+ X

+ ··· + X

where

1iftheith trial is a success

0iftheith trial is a failure

Hence, X

is a Bernoulli random variable having expectation E[X

] = 1(p) + 0(1 −

p). Thus,

E[X] = E[X

] + E[X

] + ··· + E[X

] = np .

EXAMPLE 2f Mean of a negative binomial random variable

If independent trials having a constant probability p of being successes are performed,

determine the expected number of trials required to amass a total of r successes.

Solution. If X denotes the number of trials needed to amass a total of r successes,

then X is a negative binomial random variable that can be represented by

X = X

+ X

+ ··· + X

where X

is the number of trials required to obtain the ﬁrst success, X

the number

of additional trials until the second success is obtained, X

the number of additional

302 Chapter 7 Properties of Expectation

trials until the third success is obtained, and so on. That is, X

represents the num-

ber of additional trials required after the (i − 1)st success until a total of i successes

is amassed. A little thought reveals that each of the random variables X

is a geo-

metric random variable with parameter p. Hence, from the results of Example 8b of

Chapter 4, E[X

] = 1/p, i = 1, 2, ..., r; thus,

E[X] = E[X

] + ··· + E[X

] =

EXAMPLE 2g Mean of a hypergeometric random variable

If n balls are randomly selected from an urn containing N balls of which m are white,

ﬁnd the expected number of white balls selected.

Solution. Let X denote the number of white balls selected, and represent X as

X = X

+ ··· + X

where

1iftheith white ball is selected

0 otherwise

Now

E[X

] = P{X

= 1}

= P{ith white ball is selected}





N − 1

n − 1







Hence,

E[X] = E[X

] + ··· + E[X

] =

We could also have obtained the preceding result by using the alternative represen-

tation

X = Y

+ ··· + Y

where

1iftheith ball selected is white

0 otherwise

Since the ith ball selected is equally likely to be any of the N balls, it follows that

E[Y

] =

E[X] = E[Y

] + ··· + E[Y

] =

Section 7.2 Expectation of Sums of Random Variables 303

EXAMPLE 2h Expected number of matches

Suppose that N people throw their hats into the center of a room. The hats are mixed

up, and each person randomly selects one. Find the expected number of people that

select their own hat.

Solution. Letting X denote the number of matches, we can compute E[X] most eas-

ily by writing

X = X

+ X

+ ··· + X

where

1iftheith person selects his own hat

0 otherwise

Since, for each i,theith person is equally likely to select any of the N hats,

E[X

] = P{X

= 1}=

Thus,

E[X] = E[X

] + ··· + E[X

] =





N = 1

Hence, on the average, exactly one person selects his own hat. .

EXAMPLE 2i Coupon-collecting problems

Suppose that there are N different types of coupons, and each time one obtains a

coupon, it is equally likely to be any one of the N types. Find the expected number of

coupons one need amass before obtaining a complete set of at least one of each type.

Solution. Let X denote the number of coupons collected before a complete set is

attained. We compute E[X] by using the same technique we used in computing the

mean of a negative binomial random variable (Example 2f). That is, we deﬁne X

, i =

0, 1, ..., N − 1 to be the number of additional coupons that need be obtained after

i distinct types have been collected in order to obtain another distinct type, and we

note that

X = X

+ X

+ ··· + X

N−1

When i distinct types of coupons have already been collected, a new coupon obtained

will be of a distinct type with probability (N − i)/N. Therefore,

P{X

= k}=

N − i





k−1

k Ú 1

or, in other words, X

is a geometric random variable with parameter (N − i)/N.

Hence,

E[X

] =

N − i

implying that

E[X] = 1 +

N − 1

N − 2

+ ··· +

= N



1 + ··· +

N − 1



304 Chapter 7 Properties of Expectation

EXAMPLE 2j

Ten hunters are waiting for ducks to ﬂy by. When a ﬂock of ducks ﬂies overhead, the

hunters ﬁre at the same time, but each chooses his target at random, independently

of the others. If each hunter independently hits his target with probability p, com-

pute the expected number of ducks that escape unhurt when a ﬂock of size 10 ﬂies

overhead.

Solution. Let X

equal 1 if the ith duck escapes unhurt and 0 otherwise, for i = 1,

2, ..., 10. The expected number of ducks to escape can be expressed as

E[X

+ ··· + X

] = E[X

] + ··· + E[X

]

To compute E [X

] = P{X

= 1}, we note that each of the hunters will, independently,

hit the ith duck with probability p/10, so

P{X

= 1}=



1 −



Hence,

E[X] = 10



1 −



EXAMPLE 2k Expected number of runs

Suppose that a sequence of n 1’s and m 0’s is randomly permuted so that each of the

(n + m)!/(n!m!) possible arrangements is equally likely. Any consecutive string of 1’s

is said to constitute a run of 1’s—for instance, if n = 6, m = 4, and the ordering is 1,

1, 1, 0, 1, 1, 0, 0, 1, 0, then there are 3 runs of 1’s—and we are interested in computing

the mean number of such runs. To compute this quantity, let

1 if a run of 1’s starts at the ith position

0 otherwise

Therefore, R(1), the number of runs of 1, can be expressed as

R(1) =

n+m



i=1

and it follows that

E[R(1)] =

n+m



i=1

E[I

]

Now,

E[I

] = P{“1” in position 1}

n + m

Section 7.2 Expectation of Sums of Random Variables 305

and for 1 < i … n + m,

E[I

] = P{“0” in position i − 1, “1” in position i}

n + m

n + m − 1

Hence,

E[R(1)] =

n + m

+ (n + m − 1)

(n + m)(n + m − 1)

Similarly, E[R(0)], the expected number of runs of 0’s, is

E[R(0)] =

n + m

and the expected number of runs of either type is

E[R(1) + R(0)] = 1 +

2nm

n + m

EXAMPLE 2l A random walk in the plane

Consider a particle initially located at a given point in the plane, and suppose that it

undergoes a sequence of steps of ﬁxed length, but in a completely random direction.

Speciﬁcally, suppose that the new position after each step is one unit of distance from

the previous position and at an angle of orientation from the previous position that is

uniformly distributed over (0, 2π). (See Figure 7.1.) Compute the expected square of

the distance from the origin after n steps.

␪

0 = initial position

1 = position after first step

2 = position after second step

FIGURE 7.1

Solution. Letting (X

, Y

) denote the change in position at the ith step, i = 1, ..., n,

in rectangular coordinates, we have

= cos θ

= sin θ

306 Chapter 7 Properties of Expectation

where θ

, i = 1, ..., n, are, by assumption, independent uniform (0, 2π) random vari-

ables. Because the position after n steps has rectangular coordinates





i=1



i=1



it follows that D

, the square of the distance from the origin, is given by

⎛

⎝



i=1

⎞

⎠

⎛

⎝



i=1

⎞

⎠



i=1

+ Y

) +



iZj

+ Y

)

= n +



iZj

(cos θ

cos θ

+ sin θ

sin θ

)

where cos

+ sin

= 1. Taking expectations and using the independence of θ

and

when i Z j and the fact that

2πE[cos θ

] =

2π

cos udu= sin 2π − sin 0 = 0

2πE[sin θ

] =

2π

sin udu= cos 0 − cos 2π = 0

we arrive at

E[D

] = n .

EXAMPLE 2m Analyzing the quick-sort algorithm

Suppose that we are presented with a set of n distinct values x

, x

, ..., x

and that

we desire to put them in increasing order, or as it is commonly stated, to sort them.

An efﬁcient procedure for accomplishing this task is the quick-sort algorithm, which

is deﬁned as follows. When n = 2, the algorithm compares the two values and then

puts them in the appropriate order. When n > 2, one of the elements is randomly

chosen—say it is x

—and then all of the other values are compared with x

. Those

smaller than x

are put in a bracket to the left of x

and those larger than x

are put

in a bracket to the right of x

. The algorithm then repeats itself on these brackets and

continues until all values have been sorted. For instance, suppose that we desire to

sort the following 10 distinct values:

5, 9, 3, 10, 11, 14, 8, 4, 17, 6

We start by choosing one of them at random (that is, each value has probability

being chosen). Suppose, for instance, that the value 10 is chosen. We then compare

each of the others to this value, putting in a bracket to the left of 10 all those values

smaller than 10 and to the right all those larger. This gives

{5, 9, 3, 8, 4, 6}, 10, {11, 14, 17}

We now focus on a bracketed set that contains more than a single value—say the one

on the left of the preceding—and randomly choose one of its values—say that 6 is

chosen. Comparing each of the values in the bracket with 6 and putting the smaller

Section 7.2 Expectation of Sums of Random Variables 307

ones in a new bracket to the left of 6 and the larger ones in a bracket to the right

of 6 gives

{5, 3, 4},6,{9, 8}, 10, {11, 14, 17}

If we now consider the leftmost bracket, and randomly choose the value 4 for com-

parison then the next iteration yields

{3},4,{5},6,{9, 8}, 10, {11, 14, 17}

This continues until there is no bracketed set that contains more than a single value.

If we let X denote the number of comparisons that it takes the quick-sort algorithm

to sort n distinct numbers, then E[X] is a measure of the effectiveness of this algo-

rithm. To compute E[X], we will ﬁrst express X as a sum of other random variables

as follows. To begin, give the following names to the values that are to be sorted:

Let 1 stand for the smallest, let 2 stand for the next smallest, and so on. Then, for

1 … i < j … n, let I(i, j) equal 1 if i and j are ever directly compared, and let it equal

0 otherwise. With this deﬁnition, it follows that

X =

n−1



i=1



j=i+1

I(i, j)

implying that

E[X] = E

⎡

⎢

⎣

n−1



i=1



j=i+1

I(i, j)

⎤

⎥

⎦

n−1



i=1



j=i+1

E[I(i, j)]

n−1



i=1



j=i+1

P{i and j are ever compared}

To determine the probability that i and j are ever compared, note that the values

i, i + 1, ..., j − 1, j will initially be in the same bracket (since all values are initially

in the same bracket) and will remain in the same bracket if the number chosen for

the ﬁrst comparison is not between i and j. For instance, if the comparison number is

larger than j , then all the values i, i + 1, ..., j − 1, j will go in a bracket to the left of

the comparison number, and if it is smaller than i, then they will all go in a bracket

to the right. Thus all the values i, i + 1, ..., j − 1, j will remain in the same bracket

until the ﬁrst time that one of them is chosen as a comparison value. At that point all

the other values between i and j will be compared with this comparison value. Now,

if this comparison value is neither i nor j, then upon comparison with it, i will go into

a left bracket and j into a right bracket, and thus i and j will be in different brackets

and so will never be compared. On the other hand, if the comparison value of the set

i, i + 1, ..., j − 1, j is either i or j, then there will be a direct comparison between

i and j. Now, given that the comparison value is one of the values between

i and j,

it follows that it is equally likely to be any of these j − i + 1 values, and thus the

probability that it is either i or j is 2/(j − i + 1). Therefore, we can conclude that

P{i and j are ever compared}=

j − i + 1

308 Chapter 7 Properties of Expectation

and

E[X] =

n−1



i=1



j=i+1

j − i + 1

To obtain a rough approximation of the magnitude of E[X] when n is large, we can

approximate the sums by integrals. Now



j=i+1

j − i + 1

i+1

x − i + 1

= 2log(x − i + 1)

i+1

= 2log(n − i + 1) − 2log(2)

L 2log(n − i + 1)

Thus

E[X] L

n−1



i=1

2log(n − i + 1)

L 2

n−1

log(n − x + 1) dx

= 2

log(y)dy

= 2(y log(y) − y)|

L 2n log(n)

Thus we see that when n is large, the quick-sort algorithm requires, on average,

approximately 2n log(n) comparisons to sort n distinct values. .

EXAMPLE 2n The probability of a union of events

Let A

, ...A

denote events, and deﬁne the indicator variables X

, i = 1, ..., n,by

1ifA

occurs

0 otherwise

Now, note that

1 −



i=1

(1 − X

) =

1if∪ A

occurs

0 otherwise

Hence,

⎡

⎣

1 −



i=1

(1 − X

)

⎤

⎦

= P

⎛

⎝



i=1

⎞

⎠

Section 7.2 Expectation of Sums of Random Variables 309

Expanding the left side of the preceding formula yields

⎛

⎝



i=1

⎞

⎠

= E

⎡

⎢

⎣



i=1

−



i<j



i<j<k

−··· + (−1)

n+1

···X

⎤

⎦

(2.3)

However,

···X

1ifA

···A

occurs

0 otherwise

E[X

···X

] = P(A

···A

)

Thus, Equation (2.3) is just a statement of the well-known formula for the union of

events:

P(∪A

) =



P(A

) −



i<j

P(A

) +



i<j<k

P(A

)

− ··· + (−1)

n+1

P(A

···A

) .

When one is dealing with an inﬁnite collection of random variables X

, i Ú 1, each

having a ﬁnite expectation, it is not necessarily true that

⎡

⎣



i=1

⎤

⎦



i=1

E[X

] (2.4)

To determine when (2.4) is valid, we note that



i=1

= lim

n→q



i=1

. Thus,

⎡

⎣



i=1

⎤

⎦

= E

⎡

⎣

lim

n→q



i=1

⎤

⎦

= lim

n→q

⎡

⎣



i=1

⎤

⎦

= lim

n→q



i=1

E[X

]



i=1

E[X

] (2.5)

Hence, Equation (2.4) is valid whenever we are justiﬁed in interchanging the expec-

tation and limit operations in Equation (2.5). Although, in general, this interchange

is not justiﬁed, it can be shown to be valid in two important special cases:

1. The X

are all nonnegative random variables. (That is, P{X

Ú 0}=1 for all i.)



i=1

E[|X

|] < q.

310 Chapter 7 Properties of Expectation

EXAMPLE 2o

Consider any nonnegative, integer-valued random variable X. If, for each i Ú 1, we

deﬁne

1ifX Ú i

0ifX < i

then



i=1



i=1



i=X+1



i=1

1 +



i=X+1

= X

Hence, since the X

are all nonnegative, we obtain

E[X] =



i=1

E(X

)



i=1

P{X Ú i} (2.6)

a useful identity. .

EXAMPLE 2p

Suppose that n elements—call them 1, 2, ..., n—must be stored in a computer in the

form of an ordered list. Each unit of time, a request will be made for one of these

elements—i being requested, independently of the past, with probability P(i), i Ú 1,



P(i) = 1. Assuming that these probabilities are known, what ordering minimizes

the average position in the line of the element requested?

Solution. Suppose that the elements are numbered so that P(1) Ú P(2) Ú ···Ú P(n).

To show that 1, 2, ..., n is the optimal ordering, let X denote the position of the

requested element. Now, under any ordering—say, O = i

, i

, ..., i

{X Ú k}=



j=k

P(i

)



j=k

P(j)

= P

1,2, ..., n

{X Ú k}

Summing over k and using Equation (2.6) yields

[X] Ú E

1,2, ..., n

[X]

thus showing that ordering the elements in decreasing order of the probability that

they are requested minimizes the expected position of the element requested. .

Section 7.2 Expectation of Sums of Random Variables 311

∗

7.2.1 Obtaining Bounds from Expectations via the Probabilistic Method

The probabilistic method is a technique for analyzing the properties of the elements

of a set by introducing probabilities on the set and then studying an element chosen

according to those probabilities. The technique was previously seen in Example 4l of

Chapter 3, where it was used to show that a set contained an element that satisﬁed a

certain property. In this subsection, we show how it can sometimes be used to bound

complicated functions.

Let f be a function on the elements of a ﬁnite set

S, and suppose that we are

interested in

m = max

s∈S

f (s)

A useful lower bound for m can often be obtained by letting S be a random element

S for which the expected value of f (S) is computable and then noting that m Ú f (S)

implies that

m Ú E[f (S)]

with strict inequality if f (S) is not a constant random variable. That is, E[f (S)]isa

lower bound on the maximum value.

EXAMPLE 2q The maximum number of Hamiltonian paths in a tournament

A round-robin tournament of n > 2 contestants is a tournament in which each of the





pair of contestants play each other exactly once. Suppose that the players are

numbered 1, 2, 3, ..., n. The permutation i

, i

, ...i

is said to be a Hamiltonian path if

beats i

, i

beats i

, ...,andi

n−1

beats i

. A problem of some interest is to determine

the largest possible number of Hamiltonian paths.

As an illustration, suppose that there are 3 players. On the one hand, one of them

wins twice, then there is a single Hamiltonian path. (For instance, if 1 wins twice and

2 beats 3, then the only Hamiltonian path is 1, 2, 3.) On the other hand, if each of

the players wins once, than there are 3 Hamiltonian paths. (For instance, if 1 beats 2,

2 beats 3, and 3 beats 1, then 1, 2, 3; 2, 3, 1; and 3, 1, 2, are all Hamiltonians). Hence,

when n = 3, there is a maximum of 3 Hamiltonian paths.

We now show that there is an outcome of the tournament that results in more than

n!/2

n−1

Hamiltonian paths. To begin, let the outcome of the tournament specify the

result of each of the





games played, and let

S denote the set of all 2





possible

tournament outcomes. Then, with f (s) deﬁned as the number of Hamiltonian paths

that result when the outcome is s ∈

S, we are asked to show that

max

f (s) Ú

n−1

To show this, consider the randomly chosen outcome S that is obtained when the

results of the





games are independent, with each contestant being equally likely

to win each encounter. To determine E[f(S)], the expected number of Hamiltonian

paths that result from the outcome S, number the n! permutations, and, for i =

1, ..., n!, let

1, if permutation i is a Hamiltonian

0, otherwise

312 Chapter 7 Properties of Expectation

Since

f (S) =



it follows that

E[f (S)] =



E[X

]

Because, by the assumed independence of the outcomes of the games, the probability

that any speciﬁed permutation is a Hamiltonian is (1/2)

n−1

, it follows that

E[X

] = P{X

= 1}=(1/2)

n−1

Therefore,

E[f (S)] = n!(1/2)

n−1

Since f (S) is not a constant random variable, the preceding equation implies that

there is an outcome of the tournament having more than n!/2

n−1

Hamiltonian

paths. .

EXAMPLE 2r

A grove of 52 trees is arranged in a circular fashion. If 15 chipmunks live in these

trees, show that there is a group of 7 consecutive trees that together house at least 3

chipmunks.

Solution. Let the neighborhood of a tree consist of that tree along with the next six

trees visited by moving in the clockwise direction. We want to show that, for any

choice of living accommodations of the 15 chipmunks, there is a tree that has at least

3 chipmunks living in its neighborhood. To show this, choose a tree at random and

let X denote the number of chipmunks that live in its neighborhood. To determine

E[X], arbitrarily number the 15 chipmunks and for i = 1, ..., 15, let

1, if chipmunk i lives in the neighborhood of the randomly chosen tree

0, otherwise

Because

X =



i=1

we obtain that

E[X] =



i=1

E[X

]

However, because X

will equal 1 if the randomly chosen tree is any of the 7 trees

consisting of the tree in which chipmunk i lives along with its 6 neighboring trees

when moving in the counterclockwise direction,

E[X

] = P{X

= 1}=

Consequently,

E[X] =

105

> 2

showing that there exists a tree with more than 2 chipmunks living in its neigh-

borhood. .

Section 7.2 Expectation of Sums of Random Variables 313

∗

7.2.2 The Maximum–Minimums Identity

We start with an identity relating the maximum of a set of numbers to the minimums

of the subsets of these numbers.

Proposition 2.2. For arbitrary numbers x

, i = 1, ..., n,

max



−



i<j

min(x

, x

) +



i<j<k

min(x

, x

)

+ ... + (−1)

n+1

min(x

, ..., x

)

Proof. We will give a probabilistic proof of the proposition. To begin, assume that

all the x

are in the interval [0, 1]. Let U be a uniform (0, 1) random variable, and

deﬁne the events A

, i = 1, ..., n,byA

={U < x

}. That is, A

is the event that

the uniform random variable is less than x

. Because at least one of these events A

will occur if U is less than at least one of the values x

, we have that

∪

U < max

Therefore,

P(∪

) = P

U < max

= max

Also,

P(A

) = P

U < x

= x

In addition, because all of the events A

, ..., A

will occur if U is less than all the

values x

, ..., x

, we see that the intersection of these events is

...A

U < min

j=1,...r

implying that

P(A

...A

) = P

U < min

j=1,...r

= min

j=1,...r

Thus, the proposition follows from the inclusion–exclusion formula for the proba-

bility of the union of events:

P(∪

) =



P(A

) −



i<j

P(A

) +



i<j<k

P(A

)

+ ... + (−1)

n+1

P(A

...A

)

When the x

are nonnegative, but not restricted to the unit interval, let c be such

that all the x

are less than c. Then the identity holds for the values y

= x

/c,and

the desired result follows by multiplying through by c. When the x

can be negative,

let b be such that x

+ b > 0 for all i. Therefore, by the preceding,

max

+ b) =



+ b) −



i<j

min(x

+ b, x

+ b)

+ ··· + (−1)

n+1

min(x

+ b, ..., x

+ b)

314 Chapter 7 Properties of Expectation

Letting

M =



−



i<j

min(x

, x

) + ··· + (−1)

n+1

min(x

, ..., x

)

we can rewrite the foregoing identity as

max

+ b = M + b



n −





+ ··· + (−1)

n+1







But

0 = (1 − 1)

= 1 − n +





+ ··· + (−1)





The preceding two equations show that

max

= M

and the proposition is proven.

It follows from Proposition 2.2 that, for any random variables X

, ..., X

max



−



i<j

min(X

, X

) + ··· + (−1)

n+1

min(X

, ..., X

)

Taking expectations of both sides of this equality yields the following relationship

between the expected value of the maximum and those of the partial minimums:



max





E[X

] −



i<j

E[min(X

, X

)]

+ ··· + (−1)

n+1

E[min(X

, ..., X

)] (2.7)

EXAMPLE 2s Coupon collecting with unequal probabilities

Suppose there are n different types of coupons and that each time one collects a

coupon, it is, independently of previous coupons collected, a type i coupon with prob-

ability p



i=1

= 1. Find the expected number of coupons one needs to collect to

obtain a complete set of at least one of each type.

Solution. If we let X

denote the number of coupons one needs collect to obtain a

type i, then we can express X as

X = max

i=1,...,n

Because each new coupon obtained is a type i with probability p

, X

is a geometric

random variable with parameter p

. Also, because the minimum of X

and X

is the

number of coupons needed to obtain either a type i or a type j, it follows that, for

i Z j,min(X

, X

) is a geometric random variable with parameter p

+ p

. Similarly,

min (X

, X

), the number needed to obtain any of types i, j,andk, is a geometric

Section 7.3 Moments of the Number of Events that Occur 315

random variable with parameter p

+ p

, and so on. Therefore, the identity (2.7)

yields

E[X] =



−



i<j

+ p



i<j<k

+ p

+ ··· + (−1)

n+1

+ ··· + p

Noting that

−px

dx =

and using the identity

1 −



i=1

(1 − e

−p

) =



−p

−



i<j

−(p

+ ··· + (−1)

n+1

−(p

+···+p

shows, upon integrating the identity, that

E[X] =

⎛

⎝

1 −



i=1

(1 − e

−p

)

⎞

⎠

is a more useful computational form. .

7.3 MOMENTS OF THE NUMBER OF EVENTS THAT OCCUR

Many of the examples solved in the previous section were of the following form: For

given events A

, ..., A

, ﬁnd E[X], where X is the number of these events that occur.

The solution then involved deﬁning an indicator variable I

for event A

such that

1, if A

occurs

0, otherwise

Because

X =



i=1

we obtained the result

E[X] = E

⎡

⎣



i=1

⎤

⎦



i=1

E[I

] =



i=1

P(A

) (3.1)

Now suppose we are interested in the number of pairs of events that occur.

Because I

will equal 1 if both A

and A

occur, and will equal 0 otherwise, it fol-

lows that the number of pairs is equal to



i<j

. But because X is the number of

events that occur, it also follows that the number of pairs of events that occur is





Consequently,







i<j

316 Chapter 7 Properties of Expectation

where there are





terms in the summation. Taking expectations yields











i<j

E[I

] =



i<j

P(A

) (3.2)



X(X − 1)





i<j

P(A

)

giving that

E[X

] − E[X] = 2



i<j

P(A

) (3.3)

which yields E[X

], and thus Var(X) = E[X

] − (E[X])

Moreover, by considering the number of distinct subsets of k events that all occur,

we see that







<...<i

···I

Taking expectations gives the identity











<...<i

E[I

···I

] =



<...<i

P(A

···A

) (3.4)

EXAMPLE 3a Moments of binomial random variables

Consider n independent trials, with each trial being a success with probability p.Let

be the event that trial i is a success. When i Z j, P(A

) = p

. Consequently,

Equation (3.2) yields











i<j





E[X(X − 1)] = n(n − 1)p

E[X

] − E[X] = n(n − 1)p

Now, E[X] =



i=1

P(A

) = np, so, from the preceding equation

Var(X) = E[X

] − (E[X])

= n(n − 1)p

+ np − (np)

= np(1 − p)

which is in agreement with the result obtained in Section 4.6.1.

In general, because P(A

···A

) = p

, we obtain from Equation (3.4) that











<...<i





or, equivalently,

E[X(X − 1) ···(X − k + 1)] = n(n − 1) ···(n − k + 1)p

Section 7.3 Moments of the Number of Events that Occur 317

The successive values E[X

], k Ú 3, can be recursively obtained from this identity.

For instance, with k = 3, it yields

E[X(X − 1)(X − 2)] = n(n − 1)(n − 2)p

E[X

− 3X

+ 2X] = n(n − 1)(n − 2)p

E[X

] = 3E[X

] − 2E[X] + n(n − 1)(n − 2)p

= 3n(n − 1)p

+ np + n(n − 1)(n − 2)p

EXAMPLE 3b Moments of hypergeometric random variables

Suppose n balls are randomly selected from an urn containing N balls, of which m

are white. Let A

be the event that the ith ball selected is white. Then X, the number

of white balls selected, is equal to the number of the events A

, ..., A

that occur.

Because the ith ball selected is equally likely to be any of the N balls, of which m are

white, P(A

) = m/N. Consequently, Equation (3.1) gives that E[X] =



i=1

P(A

) =

nm/N. Also, since

P(A

) = P(A

)P(A

) =

m − 1

N − 1

we obtain, from Equation (3.2),that











i<j

m(m − 1)

N(N − 1)





m(m − 1)

N(N − 1)

E[X(X − 1)] = n(n − 1)

m(m − 1)

N(N − 1)

showing that

E[X

] = n(n − 1)

m(m − 1)

N(N − 1)

+ E[X]

This formula yields the variance of the hypergeometric, namely,

Var(X) = E[X

] − (E[X])

= n(n − 1)

m(m − 1)

N(N − 1)

−



(n − 1)(m − 1)

N − 1

+ 1 −



which agrees with the result obtained in Example 8j of Chapter 4.

Higher moments of X are obtained by using Equation (3.4). Because

P(A

···A

) =

m(m − 1) ···(m − k + 1)

N(N − 1) ···(N − k + 1)

318 Chapter 7 Properties of Expectation

Equation (3.4) yields













m(m − 1) ···(m − k + 1)

N(N − 1) ···(N − k + 1)

E[X(X − 1) ···(X − k + 1)]

= n(n − 1) ···(n − k + 1)

m(m − 1) ···(m − k + 1)

N(N − 1) ···(N − k + 1)

EXAMPLE 3c Moments in the match problem

For i = 1, ..., N, let A

be the event that person i selects his or her own hat in the

match problem. Then

P(A

) = P(A

)P(A

) =

N − 1

which follows because, conditional on person i selecting her own hat, the hat selected

by person j is equally likely to be any of the other N − 1 hats, of which one is his

own. Consequently, with X equal to the number of people who select their own hat,

it follows from Equation (3.2) that











i<j

N(N − 1





N(N − 1)

thus showing that

E[X(X − 1)] = 1

Therefore, E[X

] = 1 + E[X]. Because E[X] =



i=1

P(A

) = 1, we obtain that

Var(X) = E[X

] − (E[X])

= 1.

Hence, both the mean and variance of the number of matches is 1. For higher moments,

we use Equation (3.4), along with the fact that P(A

···A

) =

N(N−1)···(N−k+1)

to obtain













N(N − 1) ···(N − k + 1)

E[X(X − 1) ···(X − k + 1)] = 1 .

EXAMPLE 3d Another coupon-collecting problem

Suppose that there are N distinct types of coupons and that, independently of past

types collected, each new one obtained is type j with probability p



j=1

= 1. Find

the expected value and variance of the number of different types of coupons that

appear among the ﬁrst n collected.

Solution. We will ﬁnd it more convenient to work with the number of uncollected

types. So, let Y equal the number of different types of coupons collected, and let

X = N − Y denote the number of uncollected types. With A

deﬁned as the event

that there are no type i coupons in the collection, X is equal to the number of the

Section 7.3 Moments of the Number of Events that Occur 319

events A

, ..., A

that occur. Because the types of the successive coupons collected

are independent, and, with probability 1 − p

each new coupon is not type i, we have

P(A

) = (1 − p

)

Hence, E[X] =



i=1

(1 − p

)

, from which it follows that

E[Y] = N − E[X] = N −



i=1

(1 − p

)

Similarly, because each of the n coupons collected is neither a type i nor a type j

coupon, with probability 1 − p

− p

, we have

P(A

) = (1 − p

− p

)

, i Z j

Thus,

E[X(X − 1)] = 2



i<j

P(A

) = 2



i<j

(1 − p

− p

)

E[X

] = 2



i<j

(1 − p

− p

)

+ E[X]

Hence, we obtain

Var(Y) = Var(X)

= E[X

] − (E[X])

= 2



i<j

(1 − p

− p

)



i=1

(1 − p

)

−

⎛

⎝



i=1

(1 − p

)

⎞

⎠

In the special case where p

= 1/N, i = 1, ..., N, the preceding formula gives

E[Y] = N



1 −



1 −





and

Var(Y) = N(N − 1)



1 −



+ N



1 −



− N



1 −



EXAMPLE 3e The negative hypergeometric random variables

Suppose an urn contains n + m balls, of which n are special and m are ordinary. These

items are removed one at a time, with each new removal being equally likely to be

any of the balls that remain in the urn. The random variable Y, equal to the number

of balls that need be withdrawn until a total of r special balls have been removed,

is said to have a negative hypergeometric distribution. The negative hypergeometric

distribution bears the same relationship to the hypergeometric distribution as the

negative binomial does to the binomial. That is, in both cases, rather than considering

a random variable equal to the number of successes in a ﬁxed number of trials (as are

the binomial and hypergeometric variables), they refer to the number of trials needed

to obtain a ﬁxed number of successes.

320 Chapter 7 Properties of Expectation

To obtain the probability mass function of a negative hypergeometric random vari-

able X, note that X will equal k if both

(a) the ﬁrst k − 1 withdrawals consist of r − 1 special and k − r ordinary balls and

(b) the kth ball withdrawn is special.

Consequently,

P{X = k}=



r−1



k−r





n+m

k−1



n − r + 1

n + m − k + 1

We will not, however, utilize the preceding probability mass function to obtain the

mean and variance of Y. Rather, let us number the m ordinary balls as o

, ..., o

and then, for each i = 1, ..., n, let A

be the event that o

is withdrawn before r

special balls have been removed. Then, if X is the number of the events A

, ..., A

that occur, it follows that X is the number of ordinary balls that are withdrawn before

a total of r special balls have been removed. Consequently,

Y = r + X

showing that

E[Y] = r + E[X] = r +



i=1

P(A

)

To determine P(A

), consider the n + 1 balls consisting of o

along with the n special

balls. Of these n + 1 balls, o

is equally likely to be the ﬁrst one withdrawn, or the

second one withdrawn, ..., or the ﬁnal one withdrawn. Hence, the probability that

it is among the ﬁrst r of these to be selected (and so is removed before a total or r

special balls have been withdrawn) is

n+1

. Consequently,

P(A

) =

n + 1

and

E[Y] = r + m

n + 1

r(n + m + 1)

n + 1

Thus, for instance, the expected number of cards of a well-shufﬂed deck that would

need to be turned over until a spade appears is 1 +

= 3.786, and the expected

number of cards that would need to be turned over until an ace appears is

1 +

= 10.6.

To determine Var(Y) = Var(X), we use the identity

E[X(X − 1)] = 2



i<j

P(A

)

Now, P (A

) is the probability that both o

and o

are removed before there have

been a total of r special balls removed. So consider the n + 2 balls consisting of o

, o

and the n special balls. Because all withdrawal orderings of these balls are equally

likely, the probability that o

and o

are both among the ﬁrst r + 1ofthemtobe

removed (and so are both removed before r special balls have been withdrawn) is

P(A

) =





r−1





n+2

r+1



r(r + 1)

(n + 1)(n + 2)

Section 7.3 Moments of the Number of Events that Occur 321

Consequently,

E[X(X − 1)] = 2





r(r + 1)

(n + 1)(n + 2)

E[X

] = m(m − 1)

r(r + 1)

(n + 1)(n + 2)

+ E[X]

Because E[X] = m

n+1

, this yields

Var(Y) = Var(X) = m(m − 1)

r(r + 1)

(n + 1)(n + 2)

n + 1

−



n + 1



A little algebra now shows that

Var(Y) =

mr(n + 1 − r)(n + m + 1)

(n + 1)

(n + 2)

EXAMPLE 3f Singletons in the coupon collector’s problem

Suppose that there are n distinct types of coupons and that, independently of past

types collected, each new one obtained is equally likely to be any of the n types.

Suppose also that one continues to collect coupons until a complete set of at least

one of each type has been obtained. Find the expected value and variance of the

number of types for which exactly one coupon of that type is collected.

Solution. Let X equal the number of types for which exactly one of that type is col-

lected. Also, let T

denote the ith type of coupon to be collected, and let A

be the

event that there is only a single type T

coupon in the complete set. Because X is

equal to the number of the events A

, ..., A

that occur, we have

E[X] =



i=1

P(A

)

Now, at the moment when the ﬁrst type T

coupon is collected, there remain n − i

types that need to be collected to have a complete set. Because, starting at this moment,

each of these n − i + 1 types (the n − i not yet collected and type T

) is equally

likely to be the last of these types to be collected, it follows that the type T

will be the

last of these types (and so will be a singleton) with probability

n−i+1

. Consequently,

P(A

) =

n−i+1

, yielding

E[X] =



i=1

n − i + 1



i=1

To determine the variance of the number of singletons, let S

i, j

,fori < j, be the event

that the ﬁrst type T

coupon to be collected is still the only one of its type to have

been collected at the moment that the ﬁrst type T

coupon has been collected. Then

P(A

) = P(A

i,j

)P(S

i,j

)

Now, P(S

i,j

) is the probability that when a type T

has just been collected, of the

n − i + 1 types consisting of type T

and the n − i as yet uncollected types, a type T

is not among the ﬁrst j − i of these types to be collected. Because type T

is equally

likely to be the ﬁrst, or second, or ..., n − i + 1 of these types to be collected, we have

P(S

i,j

) = 1 −

j − i

n − i + 1

n + 1 − j

n + 1 − i

322 Chapter 7 Properties of Expectation

Now, conditional on the event S

i,j

, both A

and A

will occur if, at the time the ﬁrst

type T

coupon is collected, of the n − j + 2 types consisting of types T

, T

, and the

n − j as yet uncollected types, T

and T

are both collected after the other n − j. But

this implies that

P(A

i,j

) = 2

n − j + 2

n − j + 1

Therefore,

P(A

) =

(n + 1 − i)(n + 2 − j)

, i < j

yielding

E[X(X − 1)] = 4



i<j

(n + 1 − i)(n + 2 − j)

Consequently, using the previous result for E[X], we obtain

Var(X) = 4



i<j

(n + 1 − i)(n + 2 − j)



i=1

−

⎛

⎝



i=1

⎞

⎠

7.4 COVARIANCE, VARIANCE OF SUMS, AND CORRELATIONS

The following proposition shows that the expectation of a product of independent

random variables is equal to the product of their expectations.

Proposition 4.1. If X and Y are independent, then, for any functions h and g,

E[g(X)h(Y)] = E[g(X)]E[h(Y)]

Proof. Suppose that X and Y are jointly continuous with joint density f (x, y). Then

E[g(X)h(Y)] =

−q

g(x)h(y)f (x, y) dx dy

−q

g(x)h(y)f

(x)f

(y) dx dy

−q

h(y)f

(y)dy

−q

g(x)f

(x) dx

= E[h(Y)]E[g(X)]

The proof in the discrete case is similar.

Just as the expected value and the variance of a single random variable give us

information about that random variable, so does the covariance between two random

variables give us information about the relationship between the random variables.

Deﬁnition

The covariance between X and Y, denoted by Cov (X, Y), is deﬁned by

Cov(X, Y) = E[(X − E[X])(Y − E[Y])]

Section 7.4 Covariance, Variance of Sums, and Correlations 323

Upon expanding the right side of the preceding deﬁnition, we see that

Cov(X, Y) = E[XY − E[X]Y − XE[Y] + E[Y]E[X]]

= E[XY] − E[X]E[Y] − E[X]E[Y] + E[X]E[Y]

= E[XY] − E[X]E[Y]

Note that if X and Y are independent, then, by Proposition 4.1, Cov(X, Y) = 0.

However, the converse is not true. A simple example of two dependent random vari-

ables X and Y having zero covariance is obtained by letting X be a random variable

such that

P{X

= 0}=P{X = 1}=P{X =−1}=

and deﬁning

Y =

0ifX Z 0

1ifX = 0

Now, XY = 0, so E[XY] = 0. Also, E[X] = 0. Thus,

Cov(X, Y) = E[XY] − E[X]E[Y] = 0

However, X and Y are clearly not independent.

The following proposition lists some of the properties of covariance.

Proposition 4.2.

(i) Cov(X, Y) = Cov(Y, X)

(ii) Cov(X, X) = Var(X)

(iii) Cov(aX, Y) = a Cov(X, Y)

(iv) Cov

⎛

⎜

⎝



i=1



j=1

⎞

⎟

⎠



i=1



j=1

Cov(X

, Y

)

Proof of Proposition 4.2: Parts (i) and (ii) follow immediately from the deﬁnition

of covariance, and part (iii) is left as an exercise for the reader. To prove part (iv),

which states that the covariance operation is additive (as is the operation of taking

expectations), let μ

= E[X

]andv

= E[Y

]. Then

⎡

⎣



i=1

⎤

⎦



i=1

, E

⎡

⎢

⎣



j=1

⎤

⎥

⎦



j=1

and

Cov

⎛

⎜

⎝



i=1



j=1

⎞

⎟

⎠

= E

⎡

⎢

⎣

⎛

⎝



i=1

−



i=1

⎞

⎠

⎛

⎜

⎝



j=1

−



j=1

⎞

⎟

⎠

⎤

⎥

⎦

= E

⎡

⎢

⎣



i=1

− μ

)



j=1

− v

)

⎤

⎥

⎦

324 Chapter 7 Properties of Expectation

= E

⎡

⎢

⎣



i=1



j=1

− μ

)(Y

− v

)

⎤

⎥

⎦



i=1



j=1

E[(X

− μ

)(Y

− v

)]

where the last equality follows because the expected value of a sum of random vari-

ables is equal to the sum of the expected values.

It follows from parts (ii) and (iv) of Proposition 4.2, upon taking Y

= X

, j =

1, ..., n,that

Var

⎛

⎝



i=1

⎞

⎠

= Cov

⎛

⎜

⎝



i=1



j=1

⎞

⎟

⎠



i=1



j=1

Cov(X

, X

)



i=1

Var(X

) +



iZj

Cov(X

, X

)

Since each pair of indices i, j, i Z j, appears twice in the double summation, the pre-

ceding formula is equivalent to

Var

⎛

⎝



i=1

⎞

⎠



i=1

Var(X

) + 2



i<j

Cov(X

, X

) (4.1)

If X

, ..., X

are pairwise independent, in that X

and X

are independent for i Z j,

then Equation (4.1) reduces to

Var

⎛

⎝



i=1

⎞

⎠



i=1

Var(X

)

The following examples illustrate the use of Equation (4.1).

EXAMPLE 4a

Let X

, ..., X

be independent and identically distributed random variables having

expected value μ and variance σ

, and as in Example 2c, let X =



i=1

/n be the sam-

ple mean. The quantities X

− X, i = 1, ..., n, are called deviations, as they equal the

differences between the individual data and the sample mean. The random variable



i=1

− X)

n − 1

is called the sample variance. Find (a) Var(

X)and(b)E[S

Section 7.4 Covariance, Variance of Sums, and Correlations 325

Solution.

(a) Var(

X) =





Var

⎛

⎝



i=1

⎞

⎠







i=1

Var(X

) by independence

(b) We start with the following algebraic identity:

(n − 1)S



i=1

− μ + μ − X)



i=1

− μ)



i=1

(X − μ)

− 2(X − μ)



i=1

− μ)



i=1

− μ)

+ n(X − μ)

− 2(X − μ)n(X − μ)



i=1

− μ)

− n(X − μ)

Taking expectations of the preceding yields

(n − 1)E[S

] =



i=1

E[(X

− μ)

] − nE[(X − μ)

]

= nσ

− nVar(X)

= (n − 1)σ

where the ﬁnal equality made use of part (a) of this example and the one preceding

it made use of the result of Example 2c, namely, that E[

X] = μ. Dividing through

by n − 1 shows that the expected value of the sample variance is the distribution

variance σ

. .

Our next example presents another method for obtaining the variance of a bino-

mial random variable.

EXAMPLE 4b Variance of a binomial random variable

Compute the variance of a binomial random variable X with parameters n and p.

Solution. Since such a random variable represents the number of successes in n inde-

pendent trials when each trial has the common probability p of being a success, we

may write

X = X

+ ··· + X

where the X

are independent Bernoulli random variables such that

1iftheith trial is a success

0 otherwise

326 Chapter 7 Properties of Expectation

Hence, from Equation (4.1), we obtain

Var(X) = Var(X

) + ··· + Var(X

)

But

Var(X

) = E[X

] − (E[X

])

= E[X

] − (E[X

])

since X

= X

= p − p

Thus,

Var(X) = np(1 − p) .

EXAMPLE 4c Sampling from a ﬁnite population

Consider a set of N people, each of whom has an opinion about a certain subject that

is measured by a real number v that represents the person’s “strength of

feeling” about the subject. Let v

represent the strength of feeling of person i,

i = 1, ...N.

Suppose that the quantities v

, i = 1, ..., N, are unknown and, to gather informa-

tion, a group of n of the N people is “randomly chosen” in the sense that all of the





subsets of size n are equally likely to be chosen. These n people are then ques-

tioned and their feelings determined. If S denotes the sum of the n sampled values,

determine its mean and variance.

An important application of the preceding problem is to a forthcoming election

in which each person in the population is either for or against a certain candidate or

proposition. If we take v

to equal 1 if person i is in favor and 0 if he or she is against,

then

v =



i=1

/N represents the proportion of the population that is in favor. To

estimate

v, a random sample of n people is chosen, and these people are polled. The

proportion of those polled who are in favor—that is, S/n—is often used as an estimate

Solution. For each person i, i = 1, ..., N, deﬁne an indicator variable I

to indicate

whether or not that person is included in the sample. That is,

1 if person i is in the random sample

0 otherwise

Now, S can be expressed by

S =



i=1

E[S] =



i=1

E[I

]

Section 7.4 Covariance, Variance of Sums, and Correlations 327

Var(S) =



i=1

Var(v

) + 2



i<j

Cov(v

, v

)



i=1

Var(I

) + 2



i<j

Cov(I

, I

)

Because

E[I

] =

E[I

] =

n − 1

N − 1

it follows that

Var(I

) =



1 −



Cov(I

, I

) =

n(n − 1)

N(N − 1)

−





−n(N − n)

(N − 1)

Hence,

E[S] = n



i=1

= n

Var(S) =



N − n





i=1

−

2n(N − n)

(N − 1)



i<j

The expression for Var(S) can be simpliﬁed somewhat by using the identity

+ ··· + v

)



i=1

+ 2



i<j

. After some simpliﬁcation, we obtain

Var(S) =

n(N − n)

N − 1

⎛

⎜

⎝



i=1

−

⎞

⎟

⎠

Consider now the special case in which Np of the v’s are equal to 1 and the remain-

der equal to 0. Then, in this case, S is a hypergeometric random variable and has

mean and variance given, respectively, by

E[S] = n

v = np since v =

= p

and

Var(S) =

n(N − n)

N − 1



− p



n(N − n)

N − 1

p(1 − p)

328 Chapter 7 Properties of Expectation

The quantity S/n, equal to the proportion of those sampled which have values equal

to 1, is such that





= p

Var





N − n

n(N − 1)

p(1 − p) .

The correlation of two random variables X and Y, denoted by ρ(X, Y), is deﬁned,

as long as Var(X) Var(Y) is positive, by

ρ(X, Y) =

Cov(X, Y)

√

Var(X)Var(Y)

It can be shown that

−1 … ρ(X, Y) … 1 (4.2)

To prove Equation (4.2), suppose that X and Y have variances given by σ

and σ

respectively. Then, on the one hand,

0 … Var





Var(X)

Var(Y)

2Cov(X, Y)

= 2[1 + ρ(X, Y)]

implying that

−1 … ρ(X, Y)

On the other hand,

0 … Var



−



Var(X)

VarY

(−σ

)

−

2Cov(X, Y)

= 2[1 − ρ(X, Y)]

implying that

ρ(X, Y) … 1

which completes the proof of Equation (4.2).

In fact, since Var(Z) = 0 implies that Z is constant with probability 1 (this intu-

itive relationship will be rigorously proven in Chapter 8), it follows from the proof of

Equation (4.2) that ρ(X, Y) = 1 implies that Y = a + bX, where b = σ

/σ

> 0and

ρ(X, Y) =−1 implies that Y = a + bX, where b =−σ

/σ

< 0. We leave it as an

exercise for the reader to show that the reverse is also true: that if Y = a + bX,then

ρ(X, Y) is either +1or−1, depending on the sign of b.

The correlation coefﬁcient is a measure of the degree of linearity between X and Y.

A value of ρ(X, Y) near +1or−1 indicates a high degree of linearity between

X and Y, whereas a value near 0 indicates that such linearity is absent. A positive

Section 7.4 Covariance, Variance of Sums, and Correlations 329

value of ρ(X, Y) indicates that Y tends to increase when X does, whereas a negative

value indicates that Y tends to decrease when X increases. If ρ(X, Y) = 0, then X

and Y are said to be uncorrelated.

EXAMPLE 4d

Let I

and I

be indicator variables for the events A and B. That is,

1ifA occurs

0 otherwise

1ifB occurs

0 otherwise

Then

E[I

] = P(A)

E[I

] = P(B)

E[I

] = P(AB)

Cov(I

, I

) = P(AB) − P(A)P(B)

= P(B)[P(A|B) − P(A)]

Thus, we obtain the quite intuitive result that the indicator variables for A and B

are either positively correlated, uncorrelated, or negatively correlated, depending on

whether P(A|B) is, respectively, greater than, equal to, or less than P(A). .

Our next example shows that the sample mean and a deviation from the sample

mean are uncorrelated.

EXAMPLE 4e

Let X

, ..., X

be independent and identically distributed random variables having

variance σ

. Show that

Cov(X

− X, X) = 0

Solution. We have

Cov(X

− X, X) = Cov(X

, X) − Cov(X, X)

= Cov

⎛

⎜

⎝



j=1

⎞

⎟

⎠

− Var(



j=1

Cov(X

, X

) −

−

= 0

330 Chapter 7 Properties of Expectation

where the next-to-last equality uses the result of Example 4a and the ﬁnal equality

follows because

Cov(X

, X

) =

0ifj Z i by independence

if j = i since Var(X

) = σ

Although X and the deviation X

− X are uncorrelated, they are not, in gen-

eral, independent. However, in the special case where the X

are normal random

variables, it turns out that not only is

X independent of a single deviation, but it is

independent of the entire sequence of deviations X

− X, j = 1, ..., n. This result

will be established in Section 7.8, where we will also show that, in this case, the sam-

ple mean

X and the sample variance S

are independent, with (n − 1)S

/σ

having

a chi-squared distribution with n − 1 degrees of freedom. (See Example 4a for the

deﬁnition of S

.) .

EXAMPLE 4f

Consider m independent trials, each of which results in any of r possible outcomes

with probabilities P

, P

, ..., P



= 1. If we let N

, i = 1, ..., r, denote the num-

berofthem trials that result in outcome i,thenN

, N

, ..., N

have the multinomial

distribution

P{N

= n

, N

= n

, ..., N

= n

! ...n

···P



i=1

= m

For i Z j, it seems likely that when N

is large, N

would tend to be small; hence, it is

intuitive that they should be negatively correlated. Let us compute their covariance

by using Proposition 4.2(iv) and the representation



k=1

(k) and N



k=1

(k)

where

(k) =

1 if trial k results in outcome i

0 otherwise

(k) =

1 if trial k results in outcome j

0 otherwise

From Proposition 4.2(iv), we have

Cov(N

, N

) =



=1



k=1

Cov(I

(k), I

())

Now, on the one hand, when k Z ,

Cov(I

(k), I

()) = 0

Section 7.5 Conditional Expectation 331

since the outcome of trial k is independent of the outcome of trial . On the other hand,

Cov(I

(), I

()) = E[I

()I

()] − E[I

()]E[I

()]

= 0 − P

=−P

where the equation uses the fact that I

()I

() = 0, since trial  cannot result in both

outcome i and outcome j. Hence, we obtain

Cov(N

, N

) =−mP

which is in accord with our intuition that N

and N

are negatively correlated. .

7.5 CONDITIONAL EXPECTATION

7.5.1 Deﬁnitions

Recall that if X and Y are jointly discrete random variables, then the conditional

probability mass function of X, given that Y = y, is deﬁned, for all y such that

P{Y = y} > 0, by

X|Y

(x|y) = P{X = x|Y = y}=

p(x, y)

(y)

It is therefore natural to deﬁne, in this case, the conditional expectation of X given

that Y = y, for all values of y such that p

(y)>0, by

E[X|Y = y] =



xP{X = x|Y = y}



X|Y

(x|y)

EXAMPLE 5a

If X and Y are independent binomial random variables with identical parameters n

and p, calculate the conditional expected value of X given that X + Y = m.

Solution. Let us ﬁrst calculate the conditional probability mass function of X given

that X + Y = m.Fork … min(n, m),

P{X = k|X + Y = m}=

P{X = k, X + Y = m}

P{X + Y = m}

P{X = k, Y = m − k}

P{X + Y = m}

P{X = k}P{Y = m − k}

P{X + Y = m}





(1 − p)

n−k



m − k



m−k

(1 − p)

n−m+k





(1 − p)

2n−m





m − k







332 Chapter 7 Properties of Expectation

where we have used the fact (see Example 3f of Chapter 6) that X + Y is a binomial

random variable with parameters 2n and p. Hence, the conditional distribution of X,

given that X + Y = m, is the hypergeometric distribution, and from Example 2g, we

obtain

E[X|X + Y = m] =

Similarly, let us recall that if X and Y are jointly continuous with a joint probabil-

ity density function f (x, y), then the conditional probability density of X, given that

Y = y, is deﬁned, for all values of y such that f

(y)>0, by

X|Y

(x|y) =

f (x, y)

(y)

It is natural, in this case, to deﬁne the conditional expectation of X, given that Y =

y,by

E[X|Y = y] =

−q

X|Y

(x|y) dx

provided that f

(y)>0.

EXAMPLE 5b

Suppose that the joint density of X and Y is given by

f (x, y) =

−x/y

−y

0 < x < q,0< y < q

Compute E[X|Y = y].

Solution. We start by computing the conditional density

X|Y

(x|y) =

f (x, y)

(y)

f (x, y)

−q

f (x, y) dx

(1/y)e

−x/y

−y

(1/y)e

−x/y

−y

(1/y)e

−x/y

(1/y)e

−x/y

Section 7.5 Conditional Expectation 333

Hence, the conditional distribution of X, given that Y = y, is just the exponential

distribution with mean y. Thus,

E[X|Y = y] =

−x/y

dx = y .

Remark. Just as conditional probabilities satisfy all of the properties of ordinary

probabilities, so do conditional expectations satisfy the properties of ordinary expec-

tations. For instance, such formulas as

E[g(X)|Y = y] =

⎧

⎪

⎨

⎪

⎩



g(x)p

X|Y

(x|y) in the discrete case

−q

g(x)f

X|Y

(x|y) dx in the continuous case

and

⎡

⎣



i=1

|Y = y

⎤

⎦



i=1

E[X

|Y = y]

remain valid. As a matter of fact, conditional expectation given that Y = y can be

thought of as being an ordinary expectation on a reduced sample space consisting

only of outcomes for which Y = y. .

7.5.2 Computing Expectations by Conditioning

Let us denote by E[X|Y] that function of the random variable Y whose value at Y = y

is E[X|Y = y]. Note that E[X|Y] is itself a random variable. An extremely important

property of conditional expectations is given by the following proposition.

Proposition 5.1.

E[X] = E[E[X|Y]] (5.1)

If Y is a discrete random variable, then Equation (5.1) states that

E[X] =



E[X|Y = y]P{Y = y} (5.1a)

whereas if Y is continuous with density f

(y), then Equation (5.1) states

E[X] =

−q

E[X|Y = y]f

(y) dy (5.1b)

We now give a proof of Equation (5.1) in the case where X and Y are both discrete

random variables.

Proof of Equation (5.1) when X and Y Are Discrete: We must show that

E[X] =



E[X|Y = y]P{Y = y} (5.2)

334 Chapter 7 Properties of Expectation

Now, the right-hand side of Equation (5.2) can be written as



E[X|Y = y]P{Y = y}=



xP{X = x|Y = y}P{Y = y}



P{X = x, Y = y}

P{Y = y}



xP{X = x, Y = y}



P{X = x, Y = y}



xP{X = x}

= E[X]

and the result is proved.

One way to understand Equation (5.2) is to interpret it as follows: To calculate

E[X], we may take a weighted average of the conditional expected value of X given

that Y = y, each of the terms E[X|Y = y] being weighted by the probability of

the event on which it is conditioned. (Of what does this remind you?) This is an

extremely useful result that often enables us to compute expectations easily by ﬁrst

conditioning on some appropriate random variable. The following examples illustrate

its use.

EXAMPLE 5c

A miner is trapped in a mine containing 3 doors. The ﬁrst door leads to a tunnel that

will take him to safety after 3 hours of travel. The second door leads to a tunnel that

will return him to the mine after 5 hours of travel. The third door leads to a tunnel

that will return him to the mine after 7 hours. If we assume that the miner is at all

times equally likely to choose any one of the doors, what is the expected length of

time until he reaches safety?

Solution. Let X denote the amount of time (in hours) until the miner reaches safety,

and let Y denote the door he initially chooses. Now,

E[X] = E[X|Y = 1]P{Y = 1}+E[X|Y = 2]P{Y = 2}

+ E[X|Y = 3]P{Y = 3}

(E[X|Y = 1] + E[X|Y = 2] + E[X|Y = 3])

However,

E[X|Y = 1] = 3

E[X|Y = 2] = 5 + E[X] (5.3)

E[X|Y = 3] = 7 + E[X]

To understand why Equation (5.3) is correct, consider, for instance, E[X|Y = 2]

and reason as follows: If the miner chooses the second door, he spends 5 hours in

the tunnel and then returns to his cell. But once he returns to his cell, the prob-

lem is as before; thus his expected additional time until safety is just E[X]. Hence,

Section 7.5 Conditional Expectation 335

E[X|Y = 2] = 5 + E[X]. The argument behind the other equalities in Equation (5.3)

is similar. Hence,

E[X] =

(3 + 5 + E[X] + 7 + E[X])

E[X] = 15 .

EXAMPLE 5d Expectation of a sum of a random number of random variables

Suppose that the number of people entering a department store on a given day is

a random variable with mean 50. Suppose further that the amounts of money spent

by these customers are independent random variables having a common mean of $8.

Finally, suppose also that the amount of money spent by a customer is also inde-

pendent of the total number of customers who enter the store. What is the expected

amount of money spent in the store on a given day?

Solution. If we let N denote the number of customers that enter the store and X

the

amount spent by the ith such customer, then the total amount of money spent can be

expressed as



i=1

. Now,

⎡

⎣



⎤

⎦

= E

⎡

⎢

⎣

⎡

⎣



⎤

⎦

⎤

⎥

⎦

But

⎡

⎣



|N = n

⎤

⎦

= E

⎡

⎣



|N = n

⎤

⎦

= E

⎡

⎣



⎤

⎦

by the independence of the X

and N

= nE[X] where E[X] = E[X

]

which implies that

⎡

⎣



⎤

⎦

= NE[X]

Thus,

⎡

⎣



i=1

⎤

⎦

= E[NE[X]] = E[N]E[X]

Hence, in our example, the expected amount of money spent in the store is 50 * $8,

or $400. .

EXAMPLE 5e

The game of craps is begun by rolling an ordinary pair of dice. If the sum of the dice is

2, 3, or 12, the player loses. If it is 7 or 11, the player wins. If it is any other number i,

336 Chapter 7 Properties of Expectation

the player continues to roll the dice until the sum is either 7 or i. If it is 7, the player

loses; if it is i, the player wins. Let R denote the number of rolls of the dice in a game

of craps. Find

(a) E[R];

(b) E[R|player wins];

Solution. If we let P

denote the probability that the sum of the dice is i,then

= P

14−i

i − 1

, i = 2, ...,7

To compute E[R], we condition on S, the initial sum, giving

E[R] =



i=2

E[R|S = i]P

However,

E[R|S = i] =

⎧

⎨

⎩

1, if i = 2, 3, 7, 11, 12

1 +

+ P

, otherwise

The preceding equation follows because if the sum is a value i that does not end

the game, then the dice will continue to be rolled until the sum is either i or 7, and

the number of rolls until this occurs is a geometric random variable with parameter

+ P

. Therefore,

E[R] = 1 +



i=4

+ P



i=8

+ P

= 1 + 2(3/9 + 4/10 + 5/11) = 3.376

To determine E[R|win], let us start by determining p, the probability that the player

wins. Conditioning on S yields

p =



i=2

P{win|S = i}P

= P

+ P



i=4

+ P



i=8

+ P

= 0.493

where the preceding uses the fact that the probability of obtaining a sum of i before

one of 7 is P

/(P

+ P

). Now, let us determine the conditional probability mass

function of S, given that the player wins. Letting Q

= P{S = i|win}, we have

= Q

= 0, Q

= P

/p, Q

= P

Section 7.5 Conditional Expectation 337

and, for i = 4, 5, 6, 8, 9, 10,

P{S = i,win}

P{win}

P{win|S = i}

p(P

+ P

)

Now, conditioning on the initial sum gives

E[R|win] =



E[R|win, S = i]Q

However, as was noted in Example 2j of Chapter 6, given that the initial sum is i,

the number of additional rolls needed and the outcome (whether a win or a loss) are

independent. (This is easily seen by ﬁrst noting that, conditional on an initial sum

of i, the outcome is independent of the number of additional dice rolls needed and

then using the symmetry property of independence, which states that if event A is

independent of event B, then event B is independent of event A.) Therefore,

E[R|win] =



E[R|S = i]Q

= 1 +



i=4

+ P



i=8

+ P

= 2.938

Although we could determine E[R|player loses] exactly as we did E[R|player wins],

it is easier to use

E[R] = E[R|win]p + E[R|lose](1 − p)

implying that

E[R|lose] =

E[R] − E[R|win]p

1 − p

= 3.801 .

EXAMPLE 5f

As deﬁned in Example 5c of Chapter 6, the bivariate normal joint density function of

the random variables X and Y is

f (x, y) =

2πσ

1 − ρ

exp

⎧

⎪

⎨

⎪

⎩

−

2(1 − ρ

)

⎡

⎣



x − μ





y − μ



− 2ρ

(x − μ

)(y − μ

)

⎤

⎦

⎫

⎪

⎬

⎪

⎭

338 Chapter 7 Properties of Expectation

We will now show that ρ is the correlation between X and Y. As shown in Exam-

ple 5c, μ

= E[X], σ

= Var(X),andμ

= E[Y], σ

= Var(Y). Consequently,

Corr(X, Y) =

Cov(X, Y)

E[XY] − μ

To determine E[XY], we condition on Y. That is, we use the identity

E[XY] = E

E[XY|Y]

Recalling from Example 5c that the conditional distribution of X given that Y = y is

normal with mean μ

+ ρ

(y − μ

), we see that

E[XY|Y = y] = E[Xy|Y = y]

= yE[X|Y = y]

= y



+ ρ

(y − μ

)



= yμ

+ ρ

− μ

Consequently,

E[XY|Y] = Yμ

+ ρ

− μ

implying that

E[XY] = E



Yμ

+ ρ

− μ



= μ

E[Y] + ρ

E[Y

− μ

= μ

+ ρ



E[Y

] − μ



= μ

+ ρ

Var(Y)

= μ

+ ρσ

Therefore,

Corr(X, Y) =

ρσ

= ρ .

Sometimes E[X] is easy to compute, and we use the conditioning identity to com-

pute a conditional expected value. This approach is illustrated by our next example.

EXAMPLE 5g

Consider n independent trials, each of which results in one of the outcomes 1, ..., k,

with respective probabilities p

, ..., p



i=1

= 1. Let N

denote the number of

trials that result in outcome i, i = 1, ..., k.Fori Z j, ﬁnd

(a) E[N

> 0] and (b) E[N

> 1]

Section 7.5 Conditional Expectation 339

Solution. To solve (a), let

I =

0, if N

= 0

1, if N

> 0

Then

E[N

] = E[N

|I = 0]P{I = 0}+E[N

|I = 1]P{I = 1}

or, equivalently,

E[N

] = E[N

= 0]P{N

= 0}+E[N

> 0]P{N

> 0}

Now, the unconditional distribution of N

is binomial with parameters n, p

. Also,

given that N

= r, each of the n − r trials that do not result in outcome i will,

independently, result in outcome j with probability P (j|not i) =

1−p

. Consequently,

the conditional distribution of N

, given that N

= r, is binomial with parameters

n − r,

1−p

. (For a more detailed argument for this conclusion, see Example 4c of

Chapter 6.) Because P{N

= 0}=(1 − p

)

, the preceding equation yields

= n

1 − p

(1 − p

)

+ E[N

> 0](1 − (1 − p

)

giving the result

E[N

> 0] = np

1 − (1 − p

)

n−1

1 − (1 − p

)

We can solve part (b) in a similar manner. Let

J =

⎧

⎨

⎩

0, if N

= 0

1, if N

= 1

2, if N

> 1

Then

E[N

] = E[N

|J = 0]P{J = 0}+E[N

|J = 1]P{J = 1}

+ E[N

|J = 2]P{J = 2}

or, equivalently,

E[N

] = E[N

= 0]P{N

= 0}+E[N

= 1]P{N

= 1}

+ E[N

> 1]P{N

> 1}

This equation yields

= n

1 − p

(1 − p

)

+ (n − 1)

1 − p

(1 − p

)

n−1

+ E[N

> 1](1 − (1 − p

)

− np

(1 − p

)

n−1

)

giving the result

E[N

> 1] =

[1 − (1 − p

)

n−1

− (n − 1)p

(1 − p

)

n−2

]

1 − (1 − p

)

− np

(1 − p

)

n−1

It is also possible to obtain the variance of a random variable by conditioning. We

illustrate this approach by the following example.

340 Chapter 7 Properties of Expectation

EXAMPLE 5h Variance of the geometric distribution

Independent trials, each resulting in a success with probability p, are successively

performed. Let N be the time of the ﬁrst success. Find Var(N).

Solution. Let Y = 1 if the ﬁrst trial results in a success and Y = 0 otherwise. Now,

Var(N) = E[N

] − (E[N])

To calculate E[N

], we condition on Y as follows:

E[N

] = E[E[N

|Y]]

However,

E[N

|Y = 1] = 1

E[N

|Y = 0] = E[(1 + N)

]

These two equations follow because, on the one hand, if the ﬁrst trial results in a

success, then, clearly, N = 1; thus, N

= 1. On the other hand, if the ﬁrst trial results

in a failure, then the total number of trials necessary for the ﬁrst success will have the

same distribution as 1 (the ﬁrst trial that results in failure) plus the necessary number

of additional trials. Since the latter quantity has the same distribution as N, we obtain

E[N

|Y = 0] = E[(1 + N)

]. Hence,

E[N

] = E[N

|Y = 1]P{Y = 1}+E[N

|Y = 0]P{Y = 0}

= p + (1 − p)E[(1 + N)

]

= 1 + (1 − p)E[2N + N

]

However, as was shown in Example 8b of Chapter 4, E[N] = 1/p; therefore,

E[N

] = 1 +

2(1 − p)

+ (1 − p)E[N

]

E[N

] =

2 − p

Consequently,

Var(N) = E[N

] − (E[N])

2 − p

−





1 − p

EXAMPLE 5i

Consider a gambling situation in which there are r players, with player i initially hav-

ing n

units, n

> 0, i = 1,..., r. At each stage, two of the players are chosen to play

a game, with the winner of the game receiving 1 unit from the loser. Any player

whose fortune drops to 0 is eliminated, and this continues until a single player has all

n K



i=1

units, with that player designated as the victor. Assuming that the results

of successive games are independent and that each game is equally likely to be won

Section 7.5 Conditional Expectation 341

by either of its two players, ﬁnd the average number of stages until one of the players

has all n units.

Solution. To ﬁnd the expected number of stages played, suppose ﬁrst that there are

only 2 players, with players 1 and 2 initially having j and n − j units, respectively.

Let X

denote the number of stages that will be played, and let m

= E[X

]. Then, for

j = 1,...,n − 1,

= 1 + A

where A

is the additional number of stages needed beyond the ﬁrst stage. Taking

expectations gives

= 1 + E[A

]

Conditioning on the result of the ﬁrst stage then yields

= 1 + E[A

|1 wins ﬁrst stage]1/2 + E[A

|2 wins ﬁrst stage]1/2

Now, if player 1 wins at the ﬁrst stage, then the situation from that point on is exactly

the same as in a problem which supposes that player 1 starts with j + 1 and player 2

with n − ( j + 1) units. Consequently,

E[A

|1 wins ﬁrst stage] = m

j+1

and, analogously,

E[A

|2 wins ﬁrst stage] = m

j−1

Thus,

= 1 +

j+1

j−1

or, equivalently,

j+1

= 2m

− m

j−1

− 2, j = 1, ..., n − 1 (5.4)

Using that m

= 0, the preceding equation yields

= 2m

− 2

= 2m

− m

− 2 = 3m

− 6 = 3(m

− 2)

= 2m

− m

− 2 = 4m

− 12 = 4(m

− 3)

suggesting that

= i(m

− i + 1), i = 1, ..., n (5.5)

To prove the preceding equality, we use mathematical induction. Since we’ve already

shown the equation to be true for i = 1, 2, we take as the induction hypothesis that

it is true whenever i … j < n. Now we must prove that it is true for j + 1. Using

Equation (5.4) yields

j+1

= 2m

− m

j−1

− 2

= 2j(m

− j + 1) − (j − 1)(m

− j + 2) − 2 (by the induction hypothesis)

= (j + 1)m

− 2j

+ 2j + j

− 3j + 2 − 2

= (j + 1)m

− j

= (j + 1)(m

− j)

342 Chapter 7 Properties of Expectation

which completes the induction proof of (5.5). Letting i = n in (5.5), and using that

= 0, now yields that

= n − 1

which, again using (5.5), gives the result

= i(n − i)

Thus, the mean number of games played when there are only 2 players with initial

amounts i and n − i is the product of their initial amounts. Because both players play

all stages, this is also the mean number of stages involving player 1.

Now let us return to the problem involving r players with initial amounts n

, i =

1,...,r,



i=1

= n. Let X denote the number of stages needed to obtain a victor,

and let X

denote the number of stages involving player i. Now, from the point of

view of player i, starting with n

, he will continue to play stages, independently being

equally likely to win or lose each one, until his fortune is either n or 0. Thus, the

number of stages he plays is exactly the same as when he has a single opponent with

an initial fortune of n − n

. Consequently, by the preceding result it follows that

E[X

] = n

(n − n

)

⎡

⎣



i=1

⎤

⎦



i=1

(n − n

) = n

−



i=1

But because each stage involves two players,

X =



i=1

Taking expectations now yields

E[X] =

⎛

⎝

−



i=1

⎞

⎠

It is interesting to note that while our argument shows that the mean number of stages

does not depend on the manner in which the teams are selected at each stage, the

same is not true for the distribution of the number of stages. To see this, suppose

r = 3, n

= n

= 1, and n

= 2. If players 1 and 2 are chosen in the ﬁrst stage, then it

will take at least three stages to determine a winner, whereas if player 3 is in the ﬁrst

stage, then it is possible for there to be only two stages. .

In our next example, we use conditioning to verify a result previously noted in

Section 6.3.1: that the expected number of uniform (0, 1) random variables that need

to be added for their sum to exceed 1 is equal to e.

EXAMPLE 5j

Let U

, U

, ... be a sequence of independent uniform (0, 1) random variables. Find

E[N] when

N = min

⎧

⎨

⎩



i=1

> 1

⎫

⎬

⎭

Section 7.5 Conditional Expectation 343

Solution. We will ﬁnd E[N] by obtaining a more general result. For x ∈ [0, 1], let

N(x) = min

⎧

⎨

⎩



i=1

> x

⎫

⎬

⎭

and set

m(x) = E[N(x)]

That is, N(x) is the number of uniform (0, 1) random variables we must add until

their sum exceeds x,andm(x) is its expected value. We will now derive an equation

for m(x) by conditioning on U

. This gives, from Equation (5.1b),

m(x) =

E[N(x)|U

= y] dy (5.6)

Now,

E[N(x)|U

= y] =

1ify > x

1 + m(x − y) if y … x

(5.7)

The preceding formula is obviously true when y > x. It is also true when y … x, since,

if the ﬁrst uniform value is y, then, at that point, the remaining number of uniform

random variables needed is the same as if we were just starting and were going to add

uniform random variables until their sum exceeded x − y. Substituting Equation (5.7)

into Equation (5.6) gives

m(x) = 1 +

m(x − y) dy

= 1 +

m(u)du

by letting

u = x − y

Differentiating the preceding equation yields



(x) = m(x)

or, equivalently,



(x)

m(x)

= 1

Integrating this equation gives

log[m(x)] = x + c

m(x) = ke

Since m(0) = 1, it follows that k = 1, so we obtain

m(x) = e

Therefore, m(1), the expected number of uniform (0, 1) random variables that need

to be added until their sum exceeds 1, is equal to e. .

344 Chapter 7 Properties of Expectation

7.5.3 Computing Probabilities by Conditioning

Not only can we obtain expectations by ﬁrst conditioning on an appropriate random

variable, but we may also use this approach to compute probabilities. To see this, let

E denote an arbitrary event, and deﬁne the indicator random variable X by

X =

1ifE occurs

0ifE does not occur

It follows from the deﬁnition of X that

E[X] = P(E)

E[X|Y = y] = P(E|Y = y) for any random variable Y

Therefore, from Equations (5.1a) and (5.1b), we obtain

P(E) =



P(E|Y = y)P(Y = y) if Y is discrete

−q

P(E|Y = y)f

(y)dy if Y is continuous

(5.8)

Note that if Y is a discrete random variable taking on one of the values y

, ..., y

then, by deﬁning the events F

, i = 1, ..., n,byF

={Y = y

}, Equation (5.8) reduces

to the familiar equation

P(E) =



i=1

P(E|F

)P(F

)

where F

, ..., F

are mutually exclusive events whose union is the sample space.

EXAMPLE 5k The best-prize problem

Suppose that we are to be presented with n distinct prizes, in sequence. After being

presented with a prize, we must immediately decide whether to accept it or to reject

it and consider the next prize. The only information we are given when deciding

whether to accept a prize is the relative rank of that prize compared to ones already

seen. That is, for instance, when the ﬁfth prize is presented, we learn how it compares

with the four prizes we’ve already seen. Suppose that once a prize is rejected, it is

lost, and that our objective is to maximize the probability of obtaining the best prize.

Assuming that all n! orderings of the prizes are equally likely, how well can we do?

Solution. Rather surprisingly, we can do quite well. To see this, ﬁx a value k,0 …

k < n, and consider the strategy that rejects the ﬁrst k prizes and then accepts the

ﬁrst one that is better than all of those ﬁrst k.LetP

(best) denote the probability that

the best prize is selected when this strategy is employed. To compute this probability,

condition on X, the position of the best prize. This gives

(best) =



i=1

(best|X = i)P(X = i)



i=1

(best|X = i)

Section 7.5 Conditional Expectation 345

Now, on the one hand, if the overall best prize is among the ﬁrst k, then no prize is

ever selected under the strategy considered. That is,

(best|X = i) = 0ifi … k

On the other hand, if the best prize is in position i, where i > k, then the best prize

will be selected if the best of the ﬁrst i − 1 prizes is among the ﬁrst k (for then none

of the prizes in positions k + 1, k + 2, ..., i − 1 would be selected). But, conditional

on the best prize being in position i, it is easy to verify that all possible orderings of

the other prizes remain equally likely, which implies that each of the ﬁrst i − 1 prizes

is equally likely to be the best of that batch. Hence, we have

(best|X = i) = P{best of ﬁrst i − 1 is among the ﬁrst k|X = i}

i − 1

if i > k

From the preceding, we obtain

(best) =



i=k+1

i − 1

k+1

x − 1

log



n − 1



log





Now, if we consider the function

g(x) =

log





then



(x) =

log





−



(x) = 0 * log





= 1 * x =

Thus, since P

(best) L g(k), we see that the best strategy of the type considered is to

let the ﬁrst n/e prizes go by and then accept the ﬁrst one to appear that is better than

all of those. In addition, since g(n/e) = 1/e, the probability that this strategy selects

the best prize is approximately 1/e L .36788.

Remark. Most people are quite surprised by the size of the probability of obtain-

ing the best prize, thinking that this probability would be close to 0 when n is large.

However, even without going through the calculations, a little thought reveals that

the probability of obtaining the best prize can be made reasonably large. Consider

the strategy of letting half of the prizes go by and then selecting the ﬁrst one to appear

that is better than all of those. The probability that a prize is actually selected is the

probability that the overall best is among the second half, and this is

. In addition,

given that a prize is selected, at the time of selection that prize would have been

346 Chapter 7 Properties of Expectation

the best of more than n/2 prizes to have appeared and would thus have probability of

at least

of being the overall best. Hence, the strategy of letting the ﬁrst half of all

prizes go by and then accepting the ﬁrst one that is better than all of those prizes has

a probability greater than

of obtaining the best prize. .

EXAMPLE 5l

Let U be a uniform random variable on (0, 1), and suppose that the conditional dis-

tribution of X, given that U = p, is binomial with parameters n and p. Find the

probability mass function of X.

Solution. Conditioning on the value of U gives

P{X = i}=

P{X = i|U = p}f

(p) dp

P{X = i|U = p}dp

i!(n − i)!

(1 − p)

n−i

Now, it can be shown (a probabilistic proof is given in Section 6.6) that

(1 − p)

n−i

dp =

i!(n − i)!

(n + 1)!

Hence, we obtain

P{X = i}=

n + 1

i = 0, ..., n

That is, we obtain the surprising result that if a coin whose probability of coming up

heads is uniformly distributed over (0, 1) is ﬂipped n times, then the number of heads

occurring is equally likely to be any of the values 0, ..., n.

Because the preceding conditional distribution has such a nice form, it is worth try-

ing to ﬁnd another argument to enhance our intuition as to why such a result is true.

To do so, let U, U

, ..., U

be n + 1 independent uniform (0, 1) random variables,

and let X denote the number of the random variables U

, ..., U

that are smaller than

U. Since all the random variables U, U

, ..., U

have the same distribution, it follows

that U is equally likely to be the smallest, or second smallest, or largest of them; so

X is equally likely to be any of the values 0, 1, ..., n. However, given that U = p,the

number of the U

that are less than U is a binomial random variable with parameters

n and p, thus establishing our previous result. .

EXAMPLE 5m

Suppose that X and Y are independent continuous random variables having densities

and f

, respectively. Compute P{X < Y}.

Solution. Conditioning on the value of Y yields

P{X < Y}=

−q

P{X < Y|Y = y}f

(y) dy

−q

P{X < y|Y = y}f

(y) dy

Section 7.5 Conditional Expectation 347

−q

P{X < y}f

(y}dy by independence

−q

(y)f

(y) dy

where

(y) =

−q

(x) dx .

EXAMPLE 5n

Suppose that X and Y are independent continuous random variables. Find the distri-

bution of X + Y.

Solution. By conditioning on the value of Y, we obtain

P{X + Y < a}=

−q

P{X + Y < a|Y = y}f

(y) dy

−q

P{X + y < a|Y = y}f

(y) dy

−q

P{X < a − y}f

(y) dy

−q

(a − y)f

(y) dy .

7.5.4 Conditional Variance

Just as we have deﬁned the conditional expectation of X given the value of Y, we can

also deﬁne the conditional variance of X given that Y = y:

Var(X|Y) K E[(X − E[X|Y])

|Y]

That is, Var(X|Y) is equal to the (conditional) expected square of the difference

between X and its (conditional) mean when the value of Y is given. In other words,

Var(X|Y) is exactly analogous to the usual deﬁnition of variance, but now all expec-

tations are conditional on the fact that Y is known.

There is a very useful relationship between Var(X), the unconditional variance of

X, and Var(X|Y), the conditional variance of X given Y, that can often be applied to

compute Var(X). To obtain this relationship, note ﬁrst that, by the same reasoning

that yields Var(X) = E[X

] − (E[X])

, we have

Var(X|Y) = E[X

|Y] − (E[X|Y])

E[Var(X|Y)] = E[E[X

|Y]] − E[(E[X|Y])

]

= E[X

] − E[(E[X|Y])

] (5.9)

348 Chapter 7 Properties of Expectation

Also, since E[E[X|Y]] = E[X], we have

Var(E[X|Y]) = E[(E[X|Y])

] − (E[X])

(5.10)

Hence, by adding Equations (5.9) and (5.10), we arrive at the following proposition.

Proposition 5.2. The conditional variance formula

Var(X) = E[Var(X|Y)] + Var(E[X|Y])

EXAMPLE 5o

Suppose that by any time t the number of people that have arrived at a train depot is

a Poisson random variable with mean λt. If the initial train arrives at the depot at a

time (independent of when the passengers arrive) that is uniformly distributed over

(0, T), what are the mean and variance of the number of passengers who enter the

train?

Solution. For each t Ú 0, let N(t) denote the number of arrivals by t, and let Y denote

the time at which the train arrives. The random variable of interest is then N(Y).

Conditioning on Y gives

E[N(Y)|Y = t] = E[N(t)|Y = t]

= E[N(t)] by the independence of Y and N(t)

= λ

t since N(t) is Poisson with mean λt

Hence,

E[N(Y)|Y] = λY

so taking expectations gives

E[N(Y)] = λE[Y] =

λT

To obtain Var(N(Y)), we use the conditional variance formula:

Var(N(Y)|Y = t) = Var(N(t)|Y = t)

= Var(N(t)) by independence

= λt

Thus,

Var(N(Y)|Y) = λY

E[N(Y)|Y] = λY

Hence, from the conditional variance formula,

Var(N(Y)) = E[λY] + Var(λY)

= λ

+ λ

where we have used the fact that Var(Y) = T

/12. .

Section 7.6 Conditional Expectation and Prediction 349

EXAMPLE 5p Variance of a sum of a random number of random variables

Let X

, X

, ...be a sequence of independent and identically distributed random vari-

ables, and let N be a nonnegative integer-valued random variable that is independent

of the sequence X

, i Ú 1. To compute Var





i=1



, we condition on N:

⎡

⎣



i=1

⎤

⎦

= NE[X]

Var

⎛

⎝



i=1

⎞

⎠

= NVar(X)

The preceding result follows because, given N,



i=1

is just the sum of a ﬁxed num-

ber of independent random variables, so its expectation and variance are just the

sums of the individual means and variances, respectively. Hence, from the conditional

variance formula,

Var

⎛

⎝



i=1

⎞

⎠

= E[N]Var(X) + (E[X])

Var(N) .

7.6 CONDITIONAL EXPECTATION AND PREDICTION

Sometimes a situation arises in which the value of a random variable X is observed

and then, on the basis of the observed value, an attempt is made to predict the value

of a second random variable Y.Letg(X) denote the predictor; that is, if X is observed

to equal x,theng(x) is our prediction for the value of Y. Clearly, we would like to

choose g so that g(X) tends to be close to Y. One possible criterion for closeness is to

choose g so as to minimize E[(Y − g(X))

]. We now show that, under this criterion,

the best possible predictor of Y is g(X) = E[Y|X].

Proposition 6.1.

E[(Y − g(X))

] Ú E[(Y − E[Y|X])

]

Proof.

E[(Y − g(X))

|X] = E[(Y − E[Y|X] + E [Y|X] − g(X))

|X]

= E[(Y − E[Y|X])

|X]

+ E[(E[Y|X] − g(X))

|X]

+ 2E[(Y − E[Y|X])(E[Y|X] − g(X))|X] (6.1)

However, given X, E[Y|X] − g(X), being a function of X, can be treated as a

constant. Thus,

E[(Y − E[Y|X])(E[Y|X] − g(X))|X]

= (E[Y|X] − g(X))E[Y − E[Y|X]|X]

= (E[Y|X] − g (X))(E[Y|X] − E[Y|X])

= 0 (6.2)

350 Chapter 7 Properties of Expectation

Hence, from Equations (6.1) and (6.2), we obtain

E[(Y − g(X))

|X] Ú E[(Y − E[Y|X])

|X]

and the desired result follows by taking expectations of both sides of the preceding

expression.

Remark. A second, more intuitive, although less rigorous, argument verifying

Proposition 6.1 is as follows. It is straightforward to verify that E[(Y − c)

]ismini-

mized at c = E[Y]. (See Theoretical Exercise 1.) Thus, if we want to predict the value

of Y when there are no data available to use, the best possible prediction, in the

sense of minimizing the mean square error, is to predict that Y will equal its mean.

However, if the value of the random variable X is observed to be x, then the predic-

tion problem remains exactly as in the previous (no-data) case, with the exception

that all probabilities and expectations are now conditional on the event that X = x.

Hence, the best prediction in this situation is to predict that Y will equal its condi-

tional expected value given that X = x, thus establishing Proposition 6.1. .

EXAMPLE 6a

Suppose that the son of a man of height x (in inches) attains a height that is normally

distributed with mean x + 1 and variance 4. What is the best prediction of the height

at full growth of the son of a man who is 6 feet tall?

Solution. Formally, this model can be written as

Y = X + 1 + e

where e is a normal random variable, independent of X, having mean 0 and variance

4. The X and Y, of course, represent the heights of the man and his son, respectively.

The best prediction E[Y|X = 72] is thus equal to

E[Y|X = 72] = E[X + 1 + e|X = 72]

= 73 + E[e|X = 72]

= 73

+ E(e) by independence

= 73 .

EXAMPLE 6b

Suppose that if a signal value s is sent from location A, then the signal value received

at location B is normally distributed with parameters (s, 1). If S, the value of the signal

sent at A, is normally distributed with parameters (μ, σ

), what is the best estimate of

the signal sent if R, the value received at B, is equal to r?

Solution. Let us start by computing the conditional density of S given R. We have

S|R

(s|r) =

S, R

(s, r)

(r)

(s)f

R|S

(r|s)

(r)

= Ke

−(s−μ)

/2σ

−(r−s)

Section 7.6 Conditional Expectation and Prediction 351

where K does not depend on s. Now,

(s − μ)

2σ

(r − s)

= s



2σ



−



+ r



s + C

1 + σ

2σ

⎡

⎣

− 2



μ + rσ

1 + σ



⎤

⎦

+ C

1 + σ

2σ



s −

(μ + rσ

)

1 + σ



+ C

where C

and C

do not depend on s. Hence,

S|R

(s|r) = C exp

⎧

⎪

⎨

⎪

⎩

−



s −

(μ + rσ

)

1 + σ





1 + σ



⎫

⎪

⎬

⎪

⎭

where C does not depend on s. Thus, we may conclude that the conditional distribu-

tion of S, the signal sent, given that r is received, is normal with mean and variance

now given by

E[S|R = r] =

μ + rσ

1 + σ

Var(S|R = r) =

1 + σ

Consequently, from Proposition 6.1, given that the value received is r, the best esti-

mate, in the sense of minimizing the mean square error, for the signal sent is

E[S|R = r] =

1 + σ

μ +

1 + σ

Writing the conditional mean as we did previously is informative, for it shows that it

equals a weighted average of μ, the a priori expected value of the signal, and r,the

value received. The relative weights given to μ and r are in the same proportion to

each other as 1 (the conditional variance of the received signal when s is sent) is to σ

(the variance of the signal to be sent). .

EXAMPLE 6c

In digital signal processing, raw continuous analog data X must be quantized, or dis-

cretized, in order to obtain a digital representation. In order to quantize the raw

data X, an increasing set of numbers a

, i = 0, ;1, ;2, ..., such that lim

i→+q

= q and

lim

i→−q

=−q is ﬁxed, and the raw data are then quantized according to the interval

, a

i+1

] in which X lies. Let us denote by y

the discretized value when X ∈ (a

, a

i+1

and let Y denote the observed discretized value—that is,

Y = y

if a

< X … a

i+1

352 Chapter 7 Properties of Expectation

The distribution of Y is given by

P{Y = y

}=F

i+1

) − F

)

Suppose now that we want to choose the values y

, i = 0, ;1, ;2, ... so as to mini-

mize E[(X − Y)

], the expected mean square difference between the raw data and

their quantized version.

(a) Find the optimal values y

, i = 0, ;1, ....

For the optimal quantizer Y, show that

(b) E[Y] = E[X], so the mean square error quantizer preserves the input mean;

Solution. (a) For any quantizer Y, upon conditioning on the value of Y, we obtain

E[(X − Y)

] =



E[(X − y

)

< X … a

i+1

]P{a

< X … a

i+1

}

Now, if we let

I = i if a

< X … a

i+1

then

E[(X − y

)

< X … a

i+1

] = E[(X − y

)

|I = i]

and by Proposition 6.1, this quantity is minimized when

= E[X|I = i]

= E[X|a

< X … a

i+1

]

i+1

(x) dx

i+1

) − F

)

Now, since the optimal quantizer is given by Y = E[X|I], it follows that

(b) E[Y] = E[X]

(c)

Var(X) = E[Var(X|I)] + Var(E[X|I])

= E[E[(X − Y)

|I]] + Var(Y)

= E[(X − Y)

] + Var(Y) .

It sometimes happens that the joint probability distribution of X and Y is not

completely known; or if it is known, it is such that the calculation of E[Y|X = x]

is mathematically intractable. If, however, the means and variances of X and Y and

the correlation of X and Y are known, then we can at least determine the best linear

predictor of Y with respect to X.

To obtain the best linear predictor of Y with respect to X, we need to choose a and

b so as to minimize E[(Y − (a + bX))

]. Now,

E[(Y − (a + bX))

] = E[Y

− 2aY − 2bXY + a

+ 2abX + b

]

= E[Y

] − 2aE[Y] − 2bE[XY] + a

+ 2abE[X] + b

E[X

]

Section 7.6 Conditional Expectation and Prediction 353

Taking partial derivatives, we obtain

∂

∂a

E[(Y − a − bX)

] =−2E[Y] + 2a + 2bE[X]

∂

∂b

E[(Y − a − bX)

] =−2E[XY] + 2aE[X] + 2bE[X

]

(6.3)

Setting Equations (6.3) to 0 and solving for a and b yields the solutions

b =

E[XY] − E[X]E[Y]

E[X

] − (E[X])

Cov(X, Y)

= ρ

a = E[Y] − bE[X] = E[Y] −

ρσ

E[X]

(6.4)

where ρ = Correlation(X, Y), σ

= Var(Y),andσ

= Var(X). It is easy to ver-

ify that the values of a and b from Equation (6.4) minimize E[(Y − a − bX)

];

thus, the best (in the sense of mean square error) linear predictor Y with respect

to X is

ρσ

(X − μ

)

where μ

= E[Y]andμ

= E[X].

The mean square error of this predictor is given by





Y − μ

− ρ

(X − μ

)





= E

(Y − μ

)

+ ρ

(X − μ

)

− 2ρ

(Y − μ

)(X − μ

)

= σ

+ ρ

− 2ρ

= σ

(1 − ρ

) (6.5)

We note from Equation (6.5) that if ρ is near +1or−1, then the mean square error

of the best linear predictor is near zero. .

EXAMPLE 6d

An example in which the conditional expectation of Y given X is linear in X,and

hence in which the best linear predictor of Y with respect to X is the best overall

predictor, is when X and Y have a bivariate normal distribution. For, as shown in

Example 5c of Chapter 6, in that case,

E[Y|X = x] = μ

+ ρ

(x − μ

) .

354 Chapter 7 Properties of Expectation

7.7 MOMENT GENERATING FUNCTIONS

The moment generating function M(t) of the random variable X is deﬁned for all real

values of t by

M(t) = E[e

]

⎧

⎪

⎨

⎪

⎩



p(x) if X is discrete with mass function p(x)

−q

f (x) dx if X is continuous with density f (x)

We call M(t) the moment generating function because all of the moments of X can be

obtained by successively differentiating M(t) and then evaluating the result at t = 0.

For example,



(t) =

E[e

]

= E



)



(7.1)

= E[Xe

]

where we have assumed that the interchange of the differentiation and expectation

operators is legitimate. That is, we have assumed that

⎡

⎣



p(x)

⎤

⎦



p(x)]

in the discrete case and



f (x) dx



f (x)] dx

in the continuous case. This assumption can almost always be justiﬁed and, indeed, is

valid for all of the distributions considered in this book. Hence, from Equation (7.1),

evaluated at t = 0, we obtain



(0) = E[X]

Similarly,



(t) =



(t)

E[Xe

]

= E



(Xe

)



= E[X

]

Thus,



(0) = E[X

]

Section 7.7 Moment Generating Functions 355

In general, the nth derivative of M(t) is given by

(t) = E[X

] n Ú 1

implying that

(0) = E[X

] n Ú 1

We now compute M(t) for some common distributions.

EXAMPLE 7a Binomial distribution with parameters n and p

If X is a binomial random variable with parameters n and p,then

M(t) = E[e

]



k=0





(1 − p)

n−k



k=0





(pe

)

(1 − p)

n−k

= (pe

+ 1 − p)

where the last equality follows from the binomial theorem. Differentiation yields



(t) = n(pe

+ 1 − p)

n−1

Thus,

E[X] = M



(0) = np

Differentiating a second time yields



(t) = n(n − 1)(pe

+ 1 − p)

n−2

(pe

)

+ n(pe

+ 1 − p)

n−1

E[X

] = M



(0) = n(n − 1)p

+ np

The variance of X is given by

Var(X) = E[X

] − (E[X])

= n(n − 1)p

+ np − n

= np(1 − p)

verifying the result obtained previously. .

EXAMPLE 7b Poisson distribution with mean λ

If X is a Poisson random variable with parameter λ,then

M(t) = E[e

]



n=0

−λ

= e

−λ



n=0

(λe

)

356 Chapter 7 Properties of Expectation

= e

−λ

λe

= exp{λ(e

− 1)}

Differentiation yields



(t) = λe

exp{λ(e

− 1)}



(t) = (λe

)

exp{λ(e

− 1)}+λe

exp{λ(e

− 1)}

Thus,

E[X] = M



(0) = λ

E[X

] = M



(0) = λ

+ λ

Var(X) = E[X

] − (E[X])

= λ

Hence, both the mean and the variance of the Poisson random variable equal λ. .

EXAMPLE 7c Exponential distribution with parameter λ

M(t) = E[e

]

λe

−λx

= λ

−(λ−t)x

λ − t

for t <λ

We note from this derivation that, for the exponential distribution, M(t) is deﬁned

only for values of t less than λ. Differentiation of M(t) yields



(t) =

(λ − t)



(t) =

2λ

(λ − t)

Hence,

E[X] = M



(0) =

E[X

] = M



(0) =

The variance of X is given by

Var(X) = E[X

] − (E[X])

EXAMPLE 7d Normal distribution

We ﬁrst compute the moment generating function of a unit normal random variable

with parameters 0 and 1. Letting Z be such a random variable, we have

(t) = E[e

]

√

2π

−q

−x

Section 7.7 Moment Generating Functions 357

√

2π

−q

exp

−

− 2tx)

√

2π

−q

exp

−

(x − t)

= e

√

2π

−q

−(x−t)

= e

Hence, the moment generating function of the unit normal random variable Z is given

by M

(t) = e

. To obtain the moment generating function of an arbitrary normal

random variable, we recall (see Section 5.4) that X = μ + σ Z will have a normal

distribution with parameters μ and σ

whenever Z is a unit normal random variable.

Hence, the moment generating function of such a random variable is given by

(t) = E[e

]

= E[e

t(μ+σ Z)

]

= E[e

tμ

tσ Z

]

= e

tμ

E[e

tσ Z

]

= e

tμ

(tσ)

= e

tμ

(tσ)

= exp

+ μt

By differentiating, we obtain



(t) = (μ + tσ

) exp

+ μt



(t) = (μ + tσ

)

exp

+ μt

+ σ

exp

+ μt

Thus,

E[X] = M



(0) = μ

E[X

] = M



(0) = μ

+ σ

implying that

Var(X) = E[X

] − E([X])

= σ

Tables 7.1 and 7.2 give the moment generating functions for some common dis-

crete and continuous distributions.

An important property of moment generating functions is that the moment gen-

erating function of the sum of independent random variables equals the product of

the individual moment generating functions. To prove this, suppose that X and Y are

358 Chapter 7 Properties of Expectation

TABLE 7.1: DISCRETE PROBABILITY DISTRIBUTION

Moment

Probability mass generating

function, p(x) function, M(t) Mean Variance

Binomial with

parameters n, p;

0 … p … 1





(1 − p)

n−x

(pe

+ 1 − p)

np np(1 − p)

x = 0, 1, ..., n

Poisson with

parameter λ>0

−λ

exp{λ(e

− 1)} λλ

x = 0, 1, 2, ...

Geometric with

parameter

0 … p … 1

p(1 − p)

x−1

1 − (1 − p)e

1 − p

x = 1, 2, ...

Negative

binomial with

parameters r, p;

0 … p … 1



n − 1

r − 1



(1 − p)

n−r



1 − (1 − p)e



r(1 − p)

n = r, r + 1, ...

independent and have moment generating functions M

(t) and M

(t), respectively.

Then M

X+Y

(t), the moment generating function of X + Y, is given by

X+Y

(t) = E[e

t(X+Y)

]

= E[e

]

= E[e

]E[e

]

= M

(t)M

(t)

where the next-to-last equality follows from Proposition 4.1, since X and Y are inde-

pendent.

Another important result is that the moment generating function uniquely deter-

mines the distribution. That is, if M

(t) exists and is ﬁnite in some region about t = 0,

then the distribution of X is uniquely determined. For instance, if

(t) =





+ 1)

then it follows from Table 7.1 that X is a binomial random variable with parameters

10 and

EXAMPLE 7e

Suppose that the moment generating function of a random variable X is given by

M(t) = e

3(e

−1)

.WhatisP{X = 0}?

Section 7.7 Moment Generating Functions 359

TABLE 7.2: CONTINUOUS PROBABILITY DISTRIBUTION

Moment

generating

Probability mass function, f (x) function, M(t) Mean Variance

Uniform over (a, b) f (x) =

⎧

⎪

⎨

⎪

⎩

b − a

a < x < b

0 otherwise

− e

t(b − a)

a + b

(b − a)

Exponential with

parameter λ>0

f (x) =

λe

−λx

x Ú 0

0 x < 0

λ − t

Gamma with parameters

(s, λ), λ>0

f (x) =

⎧

⎪

⎨

⎪

⎩

λe

−λx

(λx)

s−1

(s)

x Ú 0

0 x < 0



λ − t



Normal with parameters

(μ, σ

)

f (x) =

√

2πσ

−(x−μ)

/2σ

− q < x < q exp

μt +

μσ

360 Chapter 7 Properties of Expectation

Solution. We see from Table 7.1 that M(t) = e

3(e

−1)

is the moment generating func-

tion of a Poisson random variable with mean 3. Hence, by the one-to-one correspon-

dence between moment generating functions and distribution functions, it follows

that X must be a Poisson random variable with mean 3. Thus, P{X = 0}=e

−3

. .

EXAMPLE 7f Sums of independent binomial random variables

If X and Y are independent binomial random variables with parameters (n, p)and

(m, p), respectively, what is the distribution of X + Y?

Solution. The moment generating function of X + Y is given by

X+Y

(t) = M

(t)M

(t) = (pe

+ 1 − p)

(pe

+ 1 − p)

= (pe

+ 1 − p)

m+n

However, (pe

+ 1 − p)

m+n

is the moment generating function of a binomial ran-

dom variable having parameters m + n and p. Thus, this must be the distribution

of X + Y. .

EXAMPLE 7g Sums of independent Poisson random variables

Calculate the distribution of X + Y when X and Y are independent Poisson random

variables with means respective λ

and λ

Solution.

X+Y

(t) = M

(t)M

(t)

= exp{λ

− 1)}exp{λ

− 1)}

= exp{(λ

+ λ

)(e

− 1)}

Hence, X + Y is Poisson distributed with mean λ

+ λ

, verifying the result given

in Example 3e of Chapter 6. .

EXAMPLE 7h Sums of independent normal random variables

Show that if X and Y are independent normal random variables with respective

parameters (μ

, σ

) and (μ

, σ

),thenX + Y is normal with mean μ

+ μ

and

variance σ

+ σ

Solution.

X+Y

(t) = M

(t)M

(t)

= exp

+ μ

exp

+ μ

= exp

(σ

+ σ

+ (μ

+ μ

which is the moment generating function of a normal random variable with mean

+ μ

and variance σ

+ σ

. The desired result then follows because the moment

generating function uniquely determines the distribution. .

Section 7.7 Moment Generating Functions 361

EXAMPLE 7i

Compute the moment generating function of a chi-squared random variable with n

degrees of freedom.

Solution. We can represent such a random variable as

+ ··· + Z

where Z

, ..., Z

are independent standard normal random variables. Let M(t) be its

moment generating function. Then, by the preceding,

M(t) = (E[e

])

where Z is a standard normal random variable. Now,

E[e

] =

√

2π

−q

−x

√

2π

−q

−x

/2σ

dx where σ

= (1 − 2t)

−1

= σ

= (1 − 2t)

−1/2

where the next-to-last equality uses the fact that the normal density with mean 0 and

variance σ

integrates to 1. Therefore,

M(t) = (1 − 2t)

−n/2

EXAMPLE 7j Moment generating function of the sum of a random number of

random variables

Let X

, X

, ...be a sequence of independent and identically distributed random vari-

ables, and let N be a nonnegative, integer-valued random variable that is independent

of the sequence X, i Ú 1. We want to compute the moment generating function of

Y =



i=1

(In Example 5d, Y was interpreted as the amount of money spent in a store on a

given day when both the amount spent by a customer and the number of customers

are random variables.)

To compute the moment generating function of Y, we ﬁrst condition on N as

follows:

⎡

⎢

⎣

exp

⎧

⎨

⎩



⎫

⎬

⎭

N = n

⎤

⎥

⎦

= E

⎡

⎢

⎣

exp

⎧

⎨

⎩



⎫

⎬

⎭

N = n

⎤

⎥

⎦

= E

⎡

⎢

⎣

exp

⎧

⎨

⎩



⎫

⎬

⎭

⎤

⎥

⎦

= [M

(t)]

362 Chapter 7 Properties of Expectation

where

(t) = E[e

]

Hence,

E[e

|N] = (M

(t))

Thus,

(t) = E[(M

(t))

]

The moments of Y can now be obtained upon differentiation, as follows:



(t) = E[N(M

(t))

N−1



(t)]

E[Y] = M



(0)

= E[N(M

(0))

N−1



(0)]

= E[NEX] (7.2)

= E[N]E[X]

verifying the result of Example 5d. (In this last set of equalities, we have used the fact

that M

(0) = E[e

] = 1.)

Also,



(t) = E[N(N − 1)(M

(t))

N−2



(t))

+ N(M

(t))

N−1



(t)]

E[Y

] = M



(0)

= E[N(N − 1)(E[X])

+ NE[X

]]

= (E[X])

(E[N

] − E[N]) + E[N]E[X

] (7.3)

= E[N](E[X

] − (E[X])

) + (E[X])

E[N

]

= E[N]Var(X) + (E[X])

E[N

]

Hence, from Equations (7.2) and (7.3), we have

Var(Y) = E[N]Var(X) + (E[X])

(E[N

] − (E[N])

)

= E[N]Var(X) + (E[X])

Var(N) .

EXAMPLE 7k

Let Y denote a uniform random variable on (0, 1), and suppose that, conditional on

Y = p, the random variable X has a binomial distribution with parameters n and p.

In Example 5k, we showed that X is equally likely to take on any of the values

0, 1, ..., n. Establish this result by using moment generating functions.

Solution. To compute the moment generating function of X, start by conditioning

on the value of Y. Using the formula for the binomial moment generating function

gives

E[e

|Y = p] = (pe

+ 1 − p)

Section 7.7 Moment Generating Functions 363

Now, Y is uniform on (0, 1), so, upon taking expectations, we obtain

E[e

] =

(pe

+ 1 − p)

− 1

dy (by the substitution y = pe

+ 1 − p)

n + 1

t(n+1)

− 1

n + 1

(1 + e

+ e

+ ··· + e

)

Because the preceding is the moment generating function of a random variable that

is equally likely to be any of the values 0, 1, ..., n, the desired result follows from the

fact that the moment generating function of a random variable uniquely determines

its distribution. .

7.7.1 Joint Moment Generating Functions

It is also possible to deﬁne the joint moment generating function of two or more

random variables. This is done as follows: For any n random variables X

, ..., X

the joint moment generating function, M(t

, ..., t

), is deﬁned, for all real values of

, ..., t

,by

M(t

, ..., t

) = E[e

+···+t

]

The individual moment generating functions can be obtained from M(t

, ..., t

) by

letting all but one of the t

’s be 0. That is,

(t) = E[e

] = M(0, ...,0, t,0,...,0)

where the t is in the ith place.

It can be proven (although the proof is too advanced for this text) that the joint

moment generating function M(t

, ..., t

) uniquely determines the joint distribution

of X

, ..., X

. This result can then be used to prove that the n random variables

, ..., X

are independent if and only if

M(t

, ..., t

) = M

) ···M

) (7.4)

For the proof in one direction, if the n random variables are independent, then

M(t

, ..., t

) = E[e

+···+t

)

]

= E[e

···e

]

= E[e

] ···E[e

] by independence

= M

) ···M

)

For the proof in the other direction, if Equation (7.4) is satisﬁed, then the joint

moment generating function M(t

, ..., t

) is the same as the joint moment generating

function of n independent random variables, the ith of which has the same distri-

bution as X

. As the joint moment generating function uniquely determines the joint

distribution, this must be the joint distribution; hence, the random variables are

independent.

364 Chapter 7 Properties of Expectation

EXAMPLE 7l

Let X and Y be independent normal random variables, each with mean μ and vari-

ance σ

. In Example 7a of Chapter 6, we showed that X + Y and X − Y are inde-

pendent. Let us now establish this result by computing their joint moment generating

function:

E[e

t(X+Y)+s(X−Y)

] = E[e

(t+s)X+(t−s)Y

]

= E[e

(t+s)X

]E[e

(t−s)Y

]

= e

μ(t+s)+σ

(t+s)

μ(t−s)+σ

(t−s)

= e

2μt+σ

But we recognize the preceding as the joint moment generating function of the sum

of a normal random variable with mean 2μ and variance 2σ

and an independent

normal random variable with mean 0 and variance 2σ

. Because the joint moment

generating function uniquely determines the joint distribution, it follows that X + Y

and X − Y are independent normal random variables. .

In the next example, we use the joint moment generating function to verify a result

that was established in Example 2b of Chapter 6.

EXAMPLE 7m

Suppose that the number of events that occur is a Poisson random variable with

mean λ and that each event is independently counted with probability p. Show that

the number of counted events and the number of uncounted events are independent

Poisson random variables with respective means λp and λ(1 − p).

Solution. Let X denote the total number of events, and let X

denote the number

of them that are counted. To compute the joint moment generating function of X

the number of events that are counted, and X − X

, the number that are uncounted,

start by conditioning on X to obtain

E[e

+t(X−X

)

|X = n] = e

E[e

(s−t)X

|X = n]

= e

(pe

s−t

+ 1 − p)

= (pe

+ (1 − p)e

)

which follows because, conditional on X = n, X

is a binomial random variable with

parameters n and p. Hence,

E[e

+t(X−X

)

|X] = (pe

+ (1 − p)e

)

Taking expectations of both sides of this equation yields

E[e

+t(X−X

)

] = E[(pe

+ (1 − p)e

)

]

Now, since X is Poisson with mean λ, it follows that E[e

] = e

λ(e

−1)

. Therefore, for

any positive value a we see (by letting a = e

)thatE[a

] = e

λ(a−1)

. Thus

E[e

+t(X−X

)

] = e

λ(pe

+(1−p)e

−1)

= e

λp(e

−1

)

λ(1−p)(e

−1)

Section 7.8 Additional Properties of Normal Random Variables 365

As the preceding is the joint moment generating function of independent Poisson

random variables with respective means λp and λ(1 − p), the result is proven. .

7.8 ADDITIONAL PROPERTIES OF NORMAL RANDOM VARIABLES

7.8.1 The Multivariate Normal Distribution

Let Z

, ..., Z

be a set of n independent unit normal random variables. If, for some

constants a

,1 … i … m,1 … j … n,andμ

,1 … i … m,

= a

+ ··· + a

+ μ

= a

+ ··· + a

+ μ

= a

+ ··· + a

+ μ

= a

+ ··· + a

+ μ

then the random variables X

, ..., X

are said to have a multivariate normal distri-

bution.

From the fact that the sum of independent normal random variables is itself a

normal random variable, it follows that each X

is a normal random variable with

mean and variance given, respectively, by

E[X

] = μ

Var(X

) =



j=1

Let us now consider

M(t

, ..., t

) = E[exp{t

+ ··· + t

}]

the joint moment generating function of X

, ..., X

. The ﬁrst thing to note is that

since



i=1

is itself a linear combination of the independent normal random vari-

ables Z

, ..., Z

, it is also normally distributed. Its mean and variance are

⎡

⎣



i=1

⎤

⎦



i=1

and

Var

⎛

⎝



i=1

⎞

⎠

= Cov

⎛

⎜

⎝



i=1



j=1

⎞

⎟

⎠



i=1



j=1

Cov(X

, X

)

366 Chapter 7 Properties of Expectation

Now, if Y is a normal random variable with mean μ and variance σ

,then

E[e

] = M

(t)|

t=1

= e

μ+σ

Thus,

M(t

, ..., t

) = exp

⎧

⎪

⎨

⎪

⎩



i=1



i=1



j=1

Cov(X

, X

)

⎫

⎪

⎬

⎪

⎭

which shows that the joint distribution of X

, ..., X

is completely determined from

a knowledge of the values of E[X

] and Cov(X

, X

), i, j = 1, ..., m.

It can be shown that when m = 2, the multivariate normal distribution reduces to

the bivariate normal.

EXAMPLE 8a

Find P(X < Y) for bivariate normal random variables X and Y having parameters

= E[X], μ

= E[Y], σ

= Var(X), σ

= Var(Y), ρ = Corr(X, Y)

Solution. Because X − Y is normal with mean

E[X − Y] = μ

− μ

and variance

Var(X − Y) = Var(X) + Var(−Y) + 2Cov(X, −Y)

= σ

+ σ

− 2ρσ

we obtain

P{X < Y}=P{X − Y < 0}

= P

⎧

⎪

⎨

⎪

⎩

X − Y − (μ

− μ

)

+ σ

− 2ρσ

−(μ

− μ

)

+ σ

− 2ρσ

⎫

⎪

⎬

⎪

⎭

= 



− μ

+ σ

− 2ρσ



EXAMPLE 8b

Suppose that the conditional distribution of X, given that  = θ , is normal with

mean θ and variance 1. Moreover, suppose that  itself is a normal random variable

with mean μ and variance σ

. Find the conditional distribution of  given that X = x.

Solution. Rather than using and then simplifying Bayes’s formula, we will solve this

problem by ﬁrst showing that X,  has a bivariate normal distribution. To do so, note

that the joint density function of X,  can be written as

X,

(x, θ) = f

X|

(x|θ)f



(θ)

where f

X|

(x|θ) is a normal density with mean θ and variance 1. However, if we let Z

be a standard normal random variable that is independent of , then the conditional

distribution of Z + , given that  = θ, is also normal with mean θ and variance 1.

Section 7.8 Additional Properties of Normal Random Variables 367

Consequently, the joint density of Z + ,  is the same as that of X, . Because the

former joint density is clearly bivariate normal (since Z +  and  are both linear

combinations of the independent normal random variables Z and ), it follows that

X,  has a bivariate normal distribution. Now,

E[X] = E[Z + ] = μ

Var(X) = Var(Z + ) = 1 + σ

and

ρ = Corr(X, )

= Corr(Z + , )

Cov(Z + , )

√

Var(Z + )Var()

1 + σ

Because X,  has a bivariate normal distribution, the conditional distribution of ,

given that X = x, is normal with mean

E[|X = x] = E[] + ρ

Var()

Var(X)

(x − E[X])

= μ +

1 + σ

(x − μ)

and variance

Var(|X = x) = Var()(1 − ρ

)

1 + σ

7.8.2 The Joint Distribution of the Sample Mean and Sample Variance

Let X

, ..., X

be independent normal random variables, each with mean μ and vari-

ance σ

.LetX =



i=1

/n denote their sample mean. Since the sum of independent

normal random variables is also a normal random variable, it follows that

X is a nor-

mal random variable with (from Examples 2c and 4a) expected value μ and variance

/n.

Now, recall from Example 4e that

Cov(

X, X

− X) = 0, i = 1, ..., n (8.1)

Also, note that since

X, X

− X, X

− X, ..., X

− X are all linear combinations

of the independent standard normals (X

− μ)/σ , i = 1, ..., n, it follows that X, X

−

X, i = 1, ..., n has a joint distribution that is multivariate normal. If we let Y be

a normal random variable, with mean μ and variance σ

/n, that is independent of

the X

, i = 1, ..., n,thenY, X

− X, i = 1, ..., n also has a multivariate normal dis-

tribution and, indeed, because of Equation (8.1), has the same expected values and

covariances as the random variables

X, X

− X, i = 1, ..., n. But since a multivariate

normal distribution is determined completely by its expected values and covariances,

368 Chapter 7 Properties of Expectation

it follows that Y, X

− X, i = 1, ..., n and X, X

− X, i = 1, ..., n have the same

joint distribution, thus showing that

X is independent of the sequence of deviations

− X, i = 1, ..., n.

Since

X is independent of the sequence of deviations X

− X, i = 1, ..., n, it is also

independent of the sample variance S



i=1

− X)

/(n − 1).

Since we already know that

X is normal with mean μ and variance σ

/n, it remains

only to determine the distribution of S

. To accomplish this, recall, from Example 4a,

the algebraic identity

(n − 1)S



i=1

− X)



i=1

− μ)

− n(X − μ)

Upon dividing the preceding equation by σ

, we obtain

(n − 1)S



X − μ

σ/

√





i=1



− μ



(8.2)

Now,



i=1



− μ



is the sum of the squares of n independent standard normal random variables and so

is a chi-squared random variable with n degrees of freedom. Hence, from Example 7i,

its moment generating function is (1 − 2t)

−n/2

. Also, because



X − μ

σ/

√



is the square of a standard normal variable, it is a chi-squared random variable with

1 degree of freedom, and so has moment generating function (1 − 2t)

−1/2

. Now, we

have seen previously that the two random variables on the left side of Equation (8.2)

are independent. Hence, as the moment generating function of the sum of indepen-

dent random variables is equal to the product of their individual moment generating

functions, we have

E[e

t(n−1)S

/σ

](1 − 2t)

−1/2

= (1 − 2t)

−n/2

E[e

t(n−1)S

/σ

] = (1 − 2t)

−(n−1)/2

But as (1 − 2t)

−(n−1)/2

is the moment generating function of a chi-squared random

variable with n − 1 degrees of freedom, we can conclude, since the moment gener-

ating function uniquely determines the distribution of the random variable, it follows

that that is the distribution of (n − 1)S

/σ

Summing up, we have shown the following.

Section 7.9 General Deﬁnition of Expectation 369

Proposition 8.1. If X

, ..., X

are independent and identically distributed normal

random variables with mean μ and variance σ

, then the sample mean X and the

sample variance S

are independent. X is a normal random variable with mean μ and

variance σ

/n;(n − 1)S

/σ

is a chi-squared random variable with n − 1 degrees of

freedom.

7.9 GENERAL DEFINITION OF EXPECTATION

Up to this point, we have deﬁned expectations only for discrete and continuous ran-

dom variables. However, there also exist random variables that are neither discrete

nor continuous, and they, too, may possess an expectation. As an example of such a

random variable, let X be a Bernoulli random variable with parameter p =

, and let

Y be a uniformly distributed random variable over the interval [0, 1]. Furthermore,

suppose that X and Y are independent, and deﬁne the new random variable W by

W =

X if X = 1

Y if X Z 1

Clearly, W is neither a discrete (since its set of possible values, [0, 1], is uncountable)

nor a continuous (since P{W = 1}=

) random variable.

In order to deﬁne the expectation of an arbitrary random variable, we require the

notion of a Stieltjes integral. Before deﬁning this integral, let us recall that, for any

function g,

g(x) dx is deﬁned by

g(x) dx = lim



i=1

g(x

)(x

− x

i−1

)

where the limit is taken over all a = x

< x

··· < x

= b as n→q and where

max

i=1,...,n

− x

i−1

)→0.

For any distribution function F, we deﬁne the Stieltjes integral of the nonnegative

function g over the interval [a, b]by

g(x) dF(x) = lim



i=1

g(x

)[F(x

) − F(x

i−1

)]

where, as before, the limit is taken over all a = x

< x

< ··· < x

= b as n→q and

where max

i=1,...,n

− x

i−1

)→0. Further, we deﬁne the Stieltjes integral over the whole

real line by

−q

g(x) dF(x) = lim

a→ − q

→ + q

g(x) dF(x)

Finally, if g is not a nonnegative function, we deﬁne g

and g

−

(x) =

g(x) if g (x) Ú 0

0ifg(x)<0

−

(x) =

0ifg(x) Ú 0

−g(x) if g (x)<0

370 Chapter 7 Properties of Expectation

Because g(x) = g

(x) − g

−

(x) and g

and g

−

are both nonnegative functions, it is

natural to deﬁne

−q

g(x) dF(x) =

−q

(x) dF(x) −

−q

−

(x) dF(x)

and we say that

−q

g(x) dF(x) exists as long as

−q

(x) dF(x) and

−q

−

(x) dF(x)

are not both equal to +q.

If X is an arbitrary random variable having cumulative distribution F, we deﬁne

the expected value of X by

E[X] =

−q

xdF(x) (9.1)

It can be shown that if X is a discrete random variable with mass function p(x),then

−q

xdF(x) =



x:p(x)>0

xp(x)

whereas if X is a continuous random variable with density function f (x),then

−q

xdF(x) =

−q

xf (x) dx

The reader should note that Equation (9.1) yields an intuitive deﬁnition of E[X];

consider the approximating sum



i=1

[F(x

) − F(x

i−1

)]

of E[X]. Because F(x

) − F(x

i−1

) is just the probability that X will be in the interval

i−1

, x

], the approximating sum multiplies the approximate value of X whenitisin

the interval (x

i−1

, x

] by the probability that it will be in that interval and then sums

over all the intervals. Clearly, as these intervals get smaller and smaller in length, we

obtain the “expected value” of X.

Stieltjes integrals are mainly of theoretical interest because they yield a compact

way of deﬁning and dealing with the properties of expectation. For instance, the use

of Stieltjes integrals avoids the necessity of having to give separate statements and

proofs of theorems for the continuous and the discrete cases. However, their prop-

erties are very much the same as those of ordinary integrals, and all of the proofs

presented in this chapter can easily be translated into proofs in the general case.

SUMMARY

If X and Y have a joint probability mass function p(x, y), then

E[g(X, Y)] =



g(x, y)p(x, y)

whereas if they have a joint density function f (x, y), then

E[g(X, Y)] =

−q

g(x, y)f (x, y) dx dy

A consequence of the preceding equations is that

E[X + Y] = E[X] + E [Y]

Summary 371

which generalizes to

⎡

⎣



i=1

⎤

⎦



i=1

E[X

]

The covariance between random variables X and Y is given by

Cov(X, Y) = E[(X − E[X])(Y − E[Y])] = E[XY] − E[X]E[Y]

A useful identity is

Cov

⎛

⎜

⎝



i=1



j=1

⎞

⎟

⎠



i=1



j=1

Cov(X

, Y

)

When n = m and Y

= X

, i = 1, ..., n, the preceding formula gives

Var

⎛

⎝



i=1

⎞

⎠



i=1

Var(X

) + 2



i<j

Cov(X

, Y

)

The correlation between X and Y, denoted by ρ(X, Y), is deﬁned by

ρ(X, Y) =

Cov(X, Y)

√

Var(X)Var(Y)

If X and Y are jointly discrete random variables, then the conditional expected value

of X, given that Y = y, is deﬁned by

E[X|Y = y] =



xP{X = x|Y = y]

If X and Y are jointly continuous random variables, then

E[X|Y = y] =

−q

X|Y

(x|y)

where

X|Y

(x|y) =

f (x, y)

(y)

is the conditional probability density of X given that Y = y. Conditional expecta-

tions, which are similar to ordinary expectations except that all probabilities are now

computed conditional on the event that Y = y, satisfy all the properties of ordinary

expectations.

Let E[X|Y] denote that function of Y whose value at Y = y is E[X|Y = y]. A very

useful identity is

E[X] = E[E[X|Y]]

In the case of discrete random variables, this equation reduces to the identity

E[X] =



E[X|Y = y]P{Y = y}

and, in the continuous case, to

E[X] =

−q

E[X|Y = y]f

(y)

372 Chapter 7 Properties of Expectation

The preceding equations can often be applied to obtain E[X] by ﬁrst “conditioning”

on the value of some other random variable Y. In addition, since, for any event A,

P(A) = E[I

], where I

is 1 if A occurs and is 0 otherwise, we can use the same

equations to compute probabilities.

The conditional variance of X, given that Y = y, is deﬁned by

Var(X|Y = y) = E[(X − E[X|Y = y])

|Y = y]

Let Var(X|Y) be that function of Y whose value at Y = y is Var(X|Y = y). The

following is known as the conditional variance formula:

Var(X) = E[Var(X|Y)] + Var(E[X|Y])

Suppose that the random variable X is to be observed and, on the basis of its value,

one must then predict the value of the random variable Y. In such a situation, it turns

out that, among all predictors, E[Y|X] has the smallest expectation of the square of

the difference between it and Y.

The moment generating function of the random variable X is deﬁned by

M(t) = E[e

]

The moments of X can be obtained by successively differentiating M(t) and then

evaluating the resulting quantity at t = 0. Speciﬁcally, we have

E[X

] =

M(t)

t=0

n = 1, 2, ...

Two useful results concerning moment generating functions are, ﬁrst, that the

moment generating function uniquely determines the distribution function of the

random variable and, second, that the moment generating function of the sum of

independent random variables is equal to the product of their moment generating

functions. These results lead to simple proofs that the sum of independent normal

(Poisson, gamma) random variables remains a normal (Poisson, gamma) random

variable.

If X

, ..., X

are all linear combinations of a ﬁnite set of independent standard

normal random variables, then they are said to have a multivariate normal distri-

bution. Their joint distribution is speciﬁed by the values of E[X

], Cov(X

, X

), i, j =

1, ..., m.

If X

, ..., X

are independent and identically distributed normal random variables,

then their sample mean

X =



i=1

and their sample variance



i=1

− X)

n − 1

are independent. The sample mean

X is a normal random variable with mean μ and

variance σ

/n; the random variable (n − 1)S

/σ

is a chi-squared random variable

with n − 1 degrees of freedom.

Problems 373

PROBLEMS

7.1. A player throws a fair die and simultaneously

ﬂips a fair coin. If the coin lands heads, then she

wins twice, and if tails, then one-half of the value

that appears on the die. Determine her expected

winnings.

7.2. The game of Clue involves 6 suspects, 6 weapons,

and 9 rooms. One of each is randomly chosen and

the object of the game is to guess the chosen three.

(a) How many solutions are possible?

In one version of the game, the selection is

made and then each of the players is randomly

given three of the remaining cards. Let S, W,

and R be, respectively, the numbers of sus-

pects, weapons, and rooms in the set of three

cards given to a speciﬁed player. Also, let X

denote the number of solutions that are possi-

ble after that player observes his or her three

cards.

(b) Express X in terms of S, W,andR.

7.3. Gambles are independent, and each one results in

the player being equally likely to win or lose 1 unit.

Let W denote the net winnings of a gambler whose

strategy is to stop gambling immediately after his

ﬁrst win. Find

(a) P{W > 0}

(b) P{W < 0}

7.4. If X and Y have joint density function

X,Y

(x, y) =

1/y,if0< y < 1, 0 < x < y

0, otherwise

ﬁnd

(a) E[XY]

(b) E[X]

7.5. The county hospital is located at the center of a

square whose sides are 3 miles wide. If an accident

occurs within this square, then the hospital sends

out an ambulance. The road network is rectangu-

lar, so the travel distance from the hospital, whose

coordinates are (0, 0), to the point (x, y)is|x|+|y|.

If an accident occurs at a point that is uniformly

distributed in the square, ﬁnd the expected travel

distance of the ambulance.

7.6. A fair die is rolled 10 times. Calculate the expected

sum of the 10 rolls.

7.7. Suppose that A and B each randomly and indepen-

dently choose 3 of 10 objects. Find the expected

number of objects

(a) chosen by both A and B;

(b) not chosen by either A or B;

7.8. N people arrive separately to a professional din-

ner. Upon arrival, each person looks to see if he

or she has any friends among those present. That

person then sits either at the table of a friend or

at an unoccupied table if none of those present

is a friend. Assuming that each of the





pairs of people is, independently, a pair of friends

with probability p, ﬁnd the expected number of

occupied tables.

Hint:Let

equal 1 or 0, depending on whether

the ith arrival sits at a previously unoccupied

table.

7.9. A total of n balls, numbered 1 through n, are put

into n urns, also numbered 1 through n in such a

way that ball i is equally likely to go into any of the

urns 1, 2, ..., i.Find

(a) the expected number of urns that are empty;

(b) the probability that none of the urns is

empty.

7.10. Consider 3 trials, each having the same proba-

bility of success. Let X denote the total num-

ber of successes in these trials. If E[X] = 1.8,

what is

(a) the largest possible value of P{X = 3}?

(b) the smallest possible value of P{X = 3}?

In both cases, construct a probability scenario that

results in P{X = 3} having the stated value.

Hint: For part (b), you might start by letting U be a

uniform random variable on (0, 1) and then deﬁn-

ing the trials in terms of the value of U.

7.11. Consider n independent ﬂips of a coin having

probability p of landing on heads. Say that a

changeover occurs whenever an outcome differs

from the one preceding it. For instance, if n =

5 and the outcome is HHTHT, then there are

3 changeovers. Find the expected number of

changeovers.

Hint: Express the number of changeovers as the

sum of n − 1 Bernoulli random variables.

7.12. A group of n men and n women is lined up at

random.

(a) Find the expected number of men who have a

woman next to them.

(b) Repeat part (a), but now assuming that the

group is randomly seated at a round table.

7.13. A set of 1000 cards numbered 1 through 1000

is randomly distributed among 1000 people with

each receiving one card. Compute the expected

number of cards that are given to people whose

age matches the number on the card.

7.14. An urn has m black balls. At each stage, a black

ball is removed and a new ball that is black with

374 Chapter 7 Properties of Expectation

probability p and white with probability 1 − p

is put in its place. Find the expected number of

stages needed until there are no more black balls

in the urn.

NOTE: The preceding has possible applications

to understanding the AIDS disease. Part of the

body’s immune system consists of a certain class

of cells known as T-cells. There are 2 types of T-

cells, called CD4 and CD8. Now, while the total

number of T-cells in AIDS sufferers is (at least in

the early stages of the disease) the same as that

in healthy individuals, it has recently been dis-

covered that the mix of CD4 and CD8 T-cells is

different. Roughly 60 percent of the T-cells of a

healthy person are of the CD4 type, whereas the

percentage of the T-cells that are of CD4 type

appears to decrease continually in AIDS sufferers.

A recent model proposes that the HIV virus (the

virus that causes AIDS) attacks CD4 cells and that

the body’s mechanism for replacing killed T-cells

does not differentiate between whether the killed

T-cell was CD4 or CD8. Instead, it just produces

a new T-cell that is CD4 with probability .6 and

CD8 with probability .4. However, although this

would seem to be a very efﬁcient way of replac-

ing killed T-cells when each one killed is equally

likely to be any of the body’s T-cells (and thus has

probability .6 of being CD4), it has dangerous con-

sequences when facing a virus that targets only the

CD4 T-cells.

7.15. In Example 2h, say that i and j, i Z j,forma

matched pair if i chooses the hat belonging to j and

j chooses the hat belonging to i. Find the expected

number of matched pairs.

7.16. Let Z be a standard normal random variable, and,

for a ﬁxed x,set

X =

Z if Z > x

0 otherwise

Show that E[X] =

√

2π

−x

7.17. Adeckofn cards numbered 1 through n is thor-

oughly shufﬂed so that all possible n! orderings can

be assumed to be equally likely. Suppose you are

to make n guesses sequentially, where the ith one

is a guess of the card in position i.LetN denote

the number of correct guesses.

(a) If you are not given any information about

your earlier guesses show that, for any strat-

egy, E[N] = 1.

(b) Suppose that after each guess you are shown

the card that was in the position in question.

What do you think is the best strategy? Show

that, under this strategy,

E[N] =

n − 1

+ ··· + 1

dx = log n

whether you are right or wrong. In this case,

it can be shown that the strategy which maxi-

mizes E[N] is one that keeps on guessing the

same card until you are told you are correct

and then changes to a new card. For this strat-

egy, show that

E[N] = 1 +

+ ··· +

L e − 1

Hint: For all parts, express N as the sum of indica-

tor (that is, Bernoulli) random variables.

7.18. Cards from an ordinary deck of 52 playing cards

are turned face up one at a time. If the 1st card

is an ace, or the 2nd a deuce, or the 3rd a three,

or ..., or the 13th a king, or the 14 an ace, and so

on, we say that a match occurs. Note that we do

not require that the (13n + 1)th card be any par-

ticular ace for a match to occur but only that it be

an ace. Compute the expected number of matches

that occur.

7.19. A certain region is inhabited by r distinct types of

a certain species of insect. Each insect caught will,

independently of the types of the previous catches,

be of type i with probability

, i = 1, ..., r



= 1

(a) Compute the mean number of insects that are

caught before the ﬁrst type 1 catch.

(b) Compute the mean number of types of insects

that are caught before the ﬁrst type 1 catch.

7.20. In an urn containing n balls, the ith ball has weight

W(i), i = 1, ..., n. The balls are removed with-

out replacement, one at a time, according to the

following rule: At each selection, the probabil-

ity that a given ball in the urn is chosen is equal

to its weight divided by the sum of the weights

remaining in the urn. For instance, if at some

time i

, ..., i

is the set of balls remaining in the

urn, then the next selection will be i

with prob-

ability W(i

)





k=1

W(i

), j = 1, ..., r. Compute

the expected number of balls that are withdrawn

before ball number 1 is removed.

7.21. For a group of 100 people, compute

(a) the expected number of days of the year that

are birthdays of exactly 3 people:

(b) the expected number of distinct birthdays.

Problems 375

7.22. How many times would you expect to roll a fair die

before all 6 sides appeared at least once?

7.23. Urn 1 contains 5 white and 6 black balls, while urn

2 contains 8 white and 10 black balls. Two balls

are randomly selected from urn 1 and are put into

urn 2. If 3 balls are then randomly selected from

urn 2, compute the expected number of white balls

in the trio.

Hint:LetX

= 1iftheith white ball initially in urn

1 is one of the three selected, and let X

= 0other-

wise. Similarly, let Y

= 1iftheith white ball from

urn 2 is one of the three selected, and let Y

= 0

otherwise. The number of white balls in the trio

can now be written as



7.24. A bottle initially contains m large pills and n small

pills. Each day, a patient randomly chooses one of

the pills. If a small pill is chosen, then that pill is

eaten. If a large pill is chosen, then the pill is bro-

ken in two; one part is returned to the bottle (and

is now considered a small pill) and the other part

is then eaten.

(a) Let X denote the number of small pills in the

bottle after the last large pill has been chosen

and its smaller half returned. Find E[X].

Hint: Deﬁne n + m indicator variables, one for

each of the small pills initially present and one

for each of the m small pills created when a large

one is split in two. Now use the argument of

Example 2m.

(b) Let Y denote the day on which the last large

pill is chosen. Find E[Y].

Hint: What is the relationship between X and Y?

7.25. Let X

, X

, ... be a sequence of independent and

identically distributed continuous random vari-

ables. Let N Ú 2 be such that

Ú X

Ú ··· Ú X

N−1

< X

That is, N is the point at which the sequence stops

decreasing. Show that E[N] = e.

Hint: First ﬁnd P{N Ú n}.

7.26. If X

, X

, ..., X

are independent and identically

distributed random variables having uniform dis-

tributions over (0, 1), ﬁnd

(a) E[max(X

, ..., X

)];

(b) E[min(X

, ..., X

)].

∗

7.27. If 101 items are distributed among 10 boxes, then

at least one of the boxes must contain more than 10

items. Use the probabilistic method to prove this

result.

∗

7.28. The k-of-r-out-of-n circular reliability system, k …

r … n, consists of n components that are arranged in

a circular fashion. Each component is either func-

tional or failed, and the system functions if there

is no block of r consecutive components of which

at least k are failed. Show that there is no way

to arrange 47 components, 8 of which are failed,

to make a functional 3-of-12-out-of-47 circular

system.

∗

7.29. There are 4 different types of coupons, the ﬁrst

2 of which compose one group and the second 2

another group. Each new coupon obtained is type i

with probability p

,wherep

= p

= 1/8, p

= 3/8. Find the expected number of coupons

that one must obtain to have at least one of

(a) all 4 types;

(b) all the types of the ﬁrst group;

(d) all the types of either group.

7.30. If X and Y are independent and identically dis-

tributed with mean μ and variance σ

, ﬁnd

E[(X − Y)

]

7.31. In Problem 6, calculate the variance of the sum of

the rolls.

7.32. In Problem 9, compute the variance of the number

of empty urns.

7.33. If E[X] = 1andVar(X) = 5, ﬁnd

(a) E[(2 + X)

];

(b) Var(4 + 3X).

7.34. If 10 married couples are randomly seated at a

round table, compute (a) the expected number and

(b) the variance of the number of wives who are

seated next to their husbands.

7.35. Cards from an ordinary deck are turned face up

one at a time. Compute the expected number of

cards that need to be turned face up in order to

obtain

(a) 2aces;

(b) 5 spades;

7.36. Let X be the number of 1’s and Y the number

of 2’s that occur in n rolls of a fair die. Compute

Cov(X, Y).

7.37. A die is rolled twice. Let X equal the sum of the

outcomes, and let Y equal the ﬁrst outcome minus

the second. Compute Cov(X, Y).

7.38. The random variables X and Y have a joint density

function given by

f (x, y) =

−2x

/x 0 … x < q,0 … y … x

0 otherwise

Compute Cov(X, Y).

7.39. Let X

, ...be independent with common mean μ

and common variance σ

,andsetY

= X

n+1

+ X

n+2

.Forj Ú 0, ﬁnd Cov(Y

, Y

n+j

376 Chapter 7 Properties of Expectation

7.40. The joint density function of X and Y is given by

f (x, y) =

−(y+x/y)

, x > 0, y > 0

Find E[X], E[Y], and show that Cov(X, Y) = 1.

7.41. A pond contains 100 ﬁsh, of which 30 are carp. If 20

ﬁsh are caught, what are the mean and variance of

the number of carp among the 20? What assump-

tions are you making?

7.42. A group of 20 people consisting of 10 men and

10 women is randomly arranged into 10 pairs of

2 each. Compute the expectation and variance of

the number of pairs that consist of a man and a

woman. Now suppose the 20 people consist of 10

married couples. Compute the mean and variance

of the number of married couples that are paired

together.

7.43. Let X

, X

, ..., X

be independent random vari-

ables having an unknown continuous distribution

function F,andletY

, Y

, ..., Y

be independent

random variables having an unknown continuous

distribution function G. Now order those n + m

variables, and let

⎧

⎨

⎩

1iftheith smallest of the n + m

variables is from the X sample

0 otherwise

The random variable R =

n+m



i=1

is the sum of the

ranks of the X sample and is the basis of a standard

statistical procedure (called the Wilcoxon sum-of-

ranks test) for testing whether F and G are iden-

tical distributions. This test accepts the hypothesis

that F = G when R is neither too large nor too

small. Assuming that the hypothesis of equality is

in fact correct, compute the mean and variance

of R.

Hint: Use the results of Example 3e.

7.44. Between two distinct methods for manufacturing

certain goods, the quality of goods produced by

method i is a continuous random variable having

distribution F

, i = 1, 2. Suppose that n goods are

produced by method 1 and m by method 2. Rank

the n + m goods according to quality, and let

⎧

⎨

⎩

1ifthejth best was produced from

method 1

2 otherwise

For the vector X

, X

, ..., X

n+m

, which consists of

n 1’s and m 2’s, let R denote the number of runs

of 1. For instance, if n = 5, m = 2, and X =

1, 2, 1, 1, 1, 1, 2, then R = 2. If F

= F

(that is,

if the two methods produce identically distributed

goods), what are the mean and variance of R?

7.45. If X

, X

,andX

are (pairwise) uncorrelated

random variables, each having mean 0 and vari-

ance 1, compute the correlations of

(a) X

+ X

and X

+ X

;

(b) X

+ X

and X

+ X

7.46. Consider the following dice game, as played at a

certain gambling casino: Players 1 and 2 roll a pair

of dice in turn. The bank then rolls the dice to

determine the outcome according to the follow-

ing rule: Player i, i = 1, 2, wins if his roll is strictly

greater than the bank’s. For i = 1, 2, let

1ifi wins

0 otherwise

and show that I

and I

are positively correlated.

Explain why this result was to be expected.

7.47. Consider a graph having n vertices labeled

1, 2, ..., n, and suppose that, between each of the





pairs of distinct vertices, an edge is indepen-

dently present with probability p. The degree of

vertex i,designatedasD

, is the number of edges

that have vertex i as one of their vertices.

(a) What is the distribution of D

(b) Find ρ(D

, D

), the correlation between D

and D

7.48. A fair die is successively rolled. Let X and Y

denote, respectively, the number of rolls necessary

to obtain a 6 and a 5. Find

(a) E[X];

(b) E[X|Y = 1];

7.49. There are two misshapen coins in a box; their

probabilities for landing on heads when they are

ﬂipped are, respectively, .4and.7. One of the coins

is to be randomly chosen and ﬂipped 10 times.

Given that two of the ﬁrst three ﬂips landed on

heads, what is the conditional expected number of

heads in the 10 ﬂips?

7.50. The joint density of X and Y is given by

f (x, y) =

−x/y

−y

,0< x < q,0< y < q

Compute E[X

|Y = y].

7.51. The joint density of X and Y is given by

f (x, y) =

−y

,0< x < y,0< y < q

Compute E[X

|Y = y].

7.52. A population is made up of r disjoint subgroups.

Let p

denote the proportion of the population that

is in subgroup i, i = 1, ..., r. If the average weight

of the members of subgroup i is w

, i = 1, ..., r,

what is the average weight of the members of the

population?

Problems 377

7.53. A prisoner is trapped in a cell containing 3 doors.

The ﬁrst door leads to a tunnel that returns him

to his cell after 2 days’ travel. The second leads

to a tunnel that returns him to his cell after 4

days’ travel. The third door leads to freedom after

1 day of travel. If it is assumed that the pris-

oner will always select doors 1, 2, and 3 with

respective probabilities .5, .3, and .2, what is the

expected number of days until the prisoner reaches

freedom?

7.54. Consider the following dice game: A pair of dice

is rolled. If the sum is 7, then the game ends and

you win 0. If the sum is not 7, then you have the

option of either stopping the game and receiv-

ing an amount equal to that sum or starting over

again. For each value of i, i = 2, ..., 12, ﬁnd your

expected return if you employ the strategy of stop-

ping the ﬁrst time that a value at least as large

as i appears. What value of i leads to the largest

expected return?

Hint:LetX

denote the return when you use the

critical value i. To compute E [X

], condition on the

initial sum.

7.55. Ten hunters are waiting for ducks to ﬂy by. When

a ﬂock of ducks ﬂies overhead, the hunters ﬁre at

the same time, but each chooses his target at ran-

dom, independently of the others. If each hunter

independently hits his target with probability .6,

compute the expected number of ducks that are

hit. Assume that the number of ducks in a ﬂock is

a Poisson random variable with mean 6.

7.56. The number of people who enter an elevator on

the ground ﬂoor is a Poisson random variable with

mean 10. If there are N ﬂoors above the ground

ﬂoor, and if each person is equally likely to get off

at any one of the N ﬂoors, independently of where

the others get off, compute the expected number

of stops that the elevator will make before dis-

charging all of its passengers.

7.57. Suppose that the expected number of accidents per

week at an industrial plant is 5. Suppose also that

the numbers of workers injured in each accident

are independent random variables with a common

mean of 2.5. If the number of workers injured in

each accident is independent of the number of

accidents that occur, compute the expected num-

ber of workers injured in a week.

7.58. A coin having probability p of coming up heads is

continually ﬂipped until both heads and tails have

appeared. Find

(a) the expected number of ﬂips;

(b) the probability that the last ﬂip lands on

heads.

7.59. There are n + 1 participants in a game. Each

person independently is a winner with probabil-

ity p. The winners share a total prize of 1 unit.

(For instance, if 4 people win, then each of them

receives

, whereas if there are no winners, then

none of the participants receive anything.) Let A

denote a speciﬁed one of the players, and let X

denote the amount that is received by A.

(a) Compute the expected total prize shared by

the players.

(b) Argue that E[X] =

1 − (1 − p)

n+1

n + 1

is a winner, and conclude that

E[(1 + B)

−1

] =

1 − (1 − p)

n+1

(n + 1)p

when B is a binomial random variable with param-

eters n and p.

7.60. Each of m + 2 players pays 1 unit to a kitty in

order to play the following game: A fair coin is to

be ﬂipped successively n times, where n is an odd

number, and the successive outcomes are noted.

Before the n ﬂips, each player writes down a pre-

diction of the outcomes. For instance, if n = 3,

then a player might write down (H, H, T),which

means that he or she predicts that the ﬁrst ﬂip

will land on heads, the second on heads, and the

third on tails. After the coins are ﬂipped, the play-

ers count their total number of correct predictions.

Thus, if the actual outcomes are all heads, then the

player who wrote (H, H, T) would have 2 correct

predictions. The total kitty of m + 2 is then evenly

split up among those players having the largest

number of correct predictions.

Since each of the coin ﬂips is equally likely to

land on either heads or tails, m of the players have

decided to make their predictions in a totally ran-

dom fashion. Speciﬁcally, they will each ﬂip one

of their own fair coins n times and then use the

result as their prediction. However, the ﬁnal 2 of

the players have formed a syndicate and will use

the following strategy: One of them will make pre-

dictions in the same random fashion as the other

m players, but the other one will then predict

exactly the opposite of the ﬁrst. That is, when the

randomizing member of the syndicate predicts an

H, the other member predicts a T.Forinstance,

if the randomizing member of the syndicate

predicts (H, H, T), then the other one predicts (T,

T, H).

(a) Argue that exactly one of the syndicate mem-

bers will have more than n/2 correct predic-

tions. (Remember, n is odd.)

(b) Let

denote the number of the m nonsyndi-

cate players that have more than n/2 correct

predictions. What is the distribution of X?

378 Chapter 7 Properties of Expectation

E[payoff to the syndicate] = (m + 2)



X + 1



(d) Use part (c) of Problem 59 to conclude that

E[payoff to the syndicate] =

2(m + 2)

m + 1



1 −





m+1



and explicitly compute this number when m =

1, 2, and 3. Because it can be shown that

2(m + 2)

m + 1



1 −





m+1



> 2

it follows that the syndicate’s strategy always

gives it a positive expected proﬁt.

7.61. Let X

, ... be independent random variables with

the common distribution function F, and sup-

pose they are independent of N, a geometric

random variable with parameter p.LetM =

max(X

, ..., X

(a) Find P{M … x} by conditioning on N.

(b) Find P{M … x|N = 1}.

(d) Use (b) and (c) to rederive the probability you

found in (a).

7.62. Let U

, U

, ... be a sequence of independent uni-

form (0, 1) random variables. In Example 5i we

showed that, for 0 … x … 1, E[N(x)] = e

,where

N(x) = min

⎧

⎨

⎩

n :



i=1

> x

⎫

⎬

⎭

This problem gives another approach to establish-

ing that result.

(a) Show by induction on n that, for 0 < x … 1

and all n Ú 0,

P{N(x) Ú n + 1}=

Hint: First condition on U

andthenusethe

induction hypothesis.

Use part (a) to conclude that

E[N(x)] = e

7.63. An urn contains 30 balls, of which 10 are red and

8 are blue. From this urn, 12 balls are randomly

withdrawn. Let X denote the number of red and Y

the number of blue balls that are withdrawn. Find

Cov(X, Y)

(a) by deﬁning appropriate indicator (that is,

Bernoulli) random variables

, Y

such that X =



i=1

, Y =



j=1

(b) by conditioning (on either X or Y) to deter-

mine E[XY].

7.64. Type i light bulbs function for a random amount

of time having mean μ

and standard deviation

, i = 1, 2. A light bulb randomly chosen from a

bin of bulbs is a type 1 bulb with probability p and

a type 2 bulb with probability 1 − p.LetX denote

the lifetime of this bulb. Find

(a) E[X];

(b) Var(X).

7.65. The number of winter storms in a good year is a

Poisson random variable with mean 3, whereas the

number in a bad year is a Poisson random variable

with mean 5. If next year will be a good year with

probability .4 or a bad year with probability .6, ﬁnd

the expected value and variance of the number of

storms that will occur.

7.66. In Example 5c, compute the variance of the length

of time until the miner reaches safety.

7.67. Consider a gambler who, at each gamble, either

wins or loses her bet with respective probabilities p

and 1 − p. A popular gambling system known

as the Kelley strategy is to always bet the frac-

tion 2p − 1 of your current fortune when p >

Compute the expected fortune after n gambles of

a gambler who starts with x units and employs the

Kelley strategy.

7.68. The number of accidents that a person has in

a given year is a Poisson random variable with

mean λ. However, suppose that the value of λ

changes from person to person, being equal to 2

for 60 percent of the population and 3 for the other

40 percent. If a person is chosen at random, what

is the probability that he will have (a) 0 accidents

and (b) exactly 3 accidents in a certain year? What

is the conditional probability that he will have 3

accidents in a given year, given that he had no acci-

dents the preceding year?

7.69. Repeat Problem 68 when the proportion of the

population having a value of λ less than x is equal

to 1 − e

−x

7.70. Consider an urn containing a large number of

coins, and suppose that each of the coins has some

probability p of turning up heads when it is ﬂipped.

However, this value of p varies from coin to coin.

Suppose that the composition of the urn is such

that if a coin is selected at random from it, then

Problems 379

the p-value of the coin can be regarded as being

the value of a random variable that is uniformly

distributed over [0, 1]. If a coin is selected at ran-

dom from the urn and ﬂipped twice, compute the

probability that

(a) the ﬁrst ﬂip results in a head;

(b) both ﬂips result in heads.

7.71. In Problem 70, suppose that the coin is tossed n

times. Let X denote the number of heads that

occur. Show that

P{X = i}=

n + 1

i = 0, 1, ..., n

Hint: Make use of the fact that

a−1

(1 − x)

b−1

dx =

(a − 1)!(b − 1)!

(a + b − 1)!

when a and b are positive integers.

7.72. Suppose that in Problem 70 we continue to ﬂip the

coin until a head appears. Let N denote the num-

ber of ﬂips needed. Find

(a) P{N Ú i}, i Ú 0;

(b) P{N = i};

7.73. In Example 6b, let S denote the signal sent and R

the signal received.

(a) Compute E[R].

(b) Compute Var(R).

(d) Compute Cov(R, S).

7.74. In Example 6c, suppose that X is uniformly dis-

tributed over (0, 1). If the discretized regions are

determined by a

= 0, a

,anda

= 1,

calculate the optimal quantizer Y and compute

E[(X − Y)

7.75. The moment generating function of X is given by

(t) = exp{2e

− 2} and that of Y by M

(t) =

(

)

.IfX and Y are independent, what are

(a) P{X + Y = 2}?

(b) P{XY = 0}?

7.76. Let X be the value of the ﬁrst die and Y the sum of

the values when two dice are rolled. Compute the

joint moment generating function of X and Y.

7.77. The joint density of X and Y is given by

f (x, y) =

√

2π

−y

−(x−y)

0 < y < q,

−q < x < q

(a) Compute the joint moment generating func-

tion of X and Y.

(b) Compute the individual moment generating

functions.

7.78. Two envelopes, each containing a check, are

placed in front of you. You are to choose one of

the envelopes, open it, and see the amount of the

check. At this point, either you can accept that

amount or you can exchange it for the check in

the unopened envelope. What should you do? Is it

possible to devise a strategy that does better than

just accepting the ﬁrst envelope?

Let A and B, A < B, denote the (unknown)

amounts of the checks, and note that the strat-

egy that randomly selects an envelope and always

accepts its check has an expected return of

(A + B)/2. Consider the following strategy: Let

F(·) be any strictly increasing (that is, continu-

ous) distribution function. Choose an envelope

randomly and open it. If the discovered check has

the value x , then accept it with probability F(x) and

exchange it with probability 1 − F(x).

(a) Show that if you employ the latter strategy,

then your expected return is greater than

(A + B)/2.

Hint: Condition on whether the ﬁrst envelope

has the value A or B.

Now consider the strategy that ﬁxes a value x

and then accepts the ﬁrst check if its value is

greater than x and exchanges it otherwise.

(b) Show that, for any x, the expected return

under the x-strategy is always at least

(A + B)/

2 and that it is strictly larger than

(A + B)/2ifx lies between A and B.

whole line, and consider the following strat-

egy: Generate the value of X,andifX = x,

then employ the x-strategy of part (b). Show

that the expected return under this strategy is

greater than (A + B)/2.

7.79. Successive weekly sales, in units of one thousand

dollars, have a bivariate normal distribution with

common mean 40, common standard deviation 6,

and correlation .6.

(a) Find the probability that the total of the next

2 weeks’ sales exceeds 90.

(b) If the correlation were .2 rather than .6, do

you think that this would increase or decrease

the answer to (a)? Explain your reasoning.

380 Chapter 7 Properties of Expectation

THEORETICAL EXERCISES

7.1. Show that E[(X − a)

] is minimized at a = E[X].

7.2. Suppose that X is a continuous random variable

with density function f . Show that E[|X − a|]is

minimized when a is equal to the median of F.

Hint:Write

E[|X − a|] =

|x − a|f (x) dx

Now break up the integral into the regions where

x < a and where x > a, and differentiate.

7.3. Prove Proposition 2.1 when

(a) X and Y have a joint probability mass func-

tion;

(b) X and Y have a joint probability density func-

tion and g(x, y) Ú 0forallx, y.

7.4. Let X be a random variable having ﬁnite expec-

tation μ and variance σ

,andletg(·) be a twice

differentiable function. Show that

E[g(X)] L g(μ) +

g”(μ)

Hint: Expand g(·) in a Taylor series about μ.Use

the ﬁrst three terms and ignore the remainder.

7.5. Let A

, A

, ..., A

be arbitrary events, and deﬁne

={at least k of the A

occur}. Show that



k=1

P(C

) =



k=1

P(A

)

Hint:LetX denote the number of the A

that

occur. Show that both sides of the preceding equa-

tion are equal to E[X].

7.6. In the text, we noted that

⎡

⎣



i=1

⎤

⎦



i=1

E[X

]

when the X

are all nonnegative random variables.

Since an integral is a limit of sums, one might

expect that



X(t)dt



E[X(t)] dt

whenever X(t),0 … t < q, are all nonnegative ran-

dom variables; and this result is indeed true. Use it

to give another proof of the result that, for a non-

negative random variable X,

E[X) =

P{X > t }dt

Hint: Deﬁne, for each nonnegative t, the random

variable X(t) by

X(t) =

1ift < X

0ift Ú X

Now relate

X(t)dt to X.

7.7. We say that X is stochastically larger than Y,writ-

ten X Ú

Y, if, for all t.

P{X > t } Ú P{Y > t}

Show that if X Ú

Y,thenE[X] Ú E[Y]when

(a) X and Y are nonnegative random variables;

(b) X and Y are arbitrary random variables.

Hint:WriteX as

X = X

− X

−

where

X if X Ú 0

0ifX < 0

, X

−

0ifX Ú 0

−X if X < 0

Similarly, represent Y as Y

− Y

−

.Thenmake

useofpart(a).

7.8. Show that X is stochastically larger than Y if and

only if

E[f (X)] Ú E[f (Y)]

for all increasing functions f .

Hint: Show that X Ú

Y,thenE[f (X)] Ú E[f (Y)]

by showing that f (X) Ú

f (Y) andthenusingThe-

oretical Exercise 7.7. To show that if E[f (X)] Ú

E[f (Y)] for all increasing functions f ,thenP{X >

t} Ú P{Y > t}, deﬁne an appropriate increasing

function f .

7.9. A coin having probability p of landing on heads

is ﬂipped n times. Compute the expected number

of runs of heads of size 1, of size 2, and of size

k,1 … k … n.

7.10. Let X

, X

, ..., X

be independent and identically

distributed positive random variables. For k … n,

ﬁnd

⎡

⎢

⎣



i=1



i=1

⎤

⎥

⎦

7.11. Consider n independent trials, each resulting in

any one of r possible outcomes with probabilities

, P

, ..., P

.LetX denote the number of out-

comes that never occur in any of the trials. Find

Theoretical Exercises 381

E[X] and show that, among all probability vectors

, ..., P

, E[X] is minimized when P

= 1/r, i =

1, ..., r.

7.12. Let X

, X

,... be a sequence of independent

random variables having the probability mass

function

P{X

= 0}=P{X

= 2}=1/2, n Ú 1

The random variable X =



n=1

is said

to have the Cantor distribution.FindE[X]and

Var(X).

7.13. Let X

, ..., X

be independent and identically dis-

tributed continuous random variables. We say that

a record value occurs at time j, j … n,ifX

Ú X

for

all 1 … i … j. Show that

(a) E[number of record values] =



j=1

1/j;

(b) Var(number of record values) =



j=1

(j − 1)/j

7.14. For Example 2i, show that the variance of the

number of coupons needed to amass a full set is

equal to

N−1



i=1

(N − i)

When N is large, this can be shown to be

approximately equal (in the sense that their ratio

approaches 1 as N

→q)toN

/6.

7.15. Consider n independent trials, the ith of which

results in a success with probability P

(a) Compute the expected number of successes in

the n trials—call it μ.

(b) For a ﬁxed value of μ, what choice of

, ..., P

maximizes the variance of the num-

ber of successes?

∗

7.16. Suppose that each of the elements of S =

{1, 2, ..., n} is to be colored either red or blue.

Show that if A

, ..., A

are subsets of S,there

is a way of doing the coloring so that at most



i=1

(1/2)

|−1

of these subsets have all their ele-

ments the same color (where |A| denotes the num-

ber of elements in the set A).

7.17. Suppose that X

and X

are independent random

variables having a common mean μ. Suppose also

that Var(X

) = σ

and Var(X

) = σ

. The value

of μ is unknown, and it is proposed that μ be esti-

mated by a weighted average of X

and X

.That

is, λX

+ (1 − λ)X

will be used as an estimate

of μ for some appropriate value of λ. Which value

of λ yields the estimate having the lowest possible

variance? Explain why it is desirable to use this

value of λ.

7.18. In Example 4f, we showed that the covariance of

the multinomial random variables N

and N

equal to −mP

by expressing N

and N

as the

sum of indicator variables. We could also have

obtained that result by using the formula

Var(N

+ N

) =Var(N

) + Var(N

) + 2Cov(N

, N

)

(a) What is the distribution of N

+ N

(b) Use the preceding identity to show that

Cov(N

, N

) =−mP

7.19. Show that X and Y are identically distributed and

not necessarily independent, then

Cov(X + Y, X − Y) = 0

7.20. The Conditional Covariance Formula. The con-

ditional covariance of X and Y, given Z,is

deﬁned by

Cov(X, Y|Z) K E[(X − E[X|Z])(Y − E[Y|Z])|Z]

(a) Show that

Cov(X, Y|Z) = E[XY|Z] − E[X|Z]E[Y|Z]

(b) Prove the conditional covariance formula

Cov(X, Y

) = E[Cov(X, Y|Z)]

+ Cov(E[X|Z], E[Y|Z])

tional variance formula.

7.21. Let X

(i)

, i = 1, ..., n, denote the order statis-

tics from a set of n uniform (0, 1) random vari-

ables, and note that the density function of X

(i)

given by

f (x) =

(i − 1)!(n − i)!

i−1

(1 − x)

n−i

0 < x < 1

(a) Compute Var(X

(i)

), i = 1, ..., n.

(b) Which value of i minimizes, and which value

maximizes, Var(X

(i)

7.22. Show that Y = a + bX,then

ρ(X, Y) =

+1ifb > 0

−1ifb < 0

7.23. Show that Z is a standard normal random variable

and if Y is deﬁned by Y = a + bZ + cZ

,then

ρ(Y, Z) =

+ 2c

382 Chapter 7 Properties of Expectation

7.24. Prove the Cauchy–Schwarz inequality, namely,

(E[XY])

… E[X

]E[Y

]

Hint:UnlessY =−tX for some constant, in which

case the inequality holds with equality, if follows

that, for all t,

0 < E[(tX + Y)

] = E[X

+ 2E[XY]t + E[Y

]

Hence, the roots of the quadratic equation

E[X

+ 2E[XY]t + E[Y

] = 0

must be imaginary, which implies that the discrim-

inant of this quadratic equation must be negative.

7.25. Show that if X and Y are independent, then

E[X|Y = y] = E[X]forally

(a) in the discrete case;

(b) in the continuous case.

7.26. Prove that E[g(X)Y|X] = g(X)E[Y|X].

7.27. Prove that if E[Y|X = x] = E[Y]forallx,thenX

and Y are uncorrelated; give a counterexample to

show that the converse is not true.

Hint: Prove and use the fact that E[XY] =

E[XE[Y|X]].

7.28. Show that Cov(X, E[Y|X])

= Cov(X, Y).

7.29. Let X

, ..., X

be independent and identically dis-

tributed random variables. Find

E[X

+ ··· + X

= x]

7.30. Consider Example 4f, which is concerned with the

multinomial distribution. Use conditional expec-

tation to compute E[N

], and then use this to

verify the formula for Cov(N

, N

) given in Exam-

ple 4f.

7.31. An urn initially contains b black and w white balls.

At each stage, we add r black balls and then with-

draw, at random, r balls from the b + w + r balls

in the urn. Show that

E[number of white balls after stage t]



b + w

b + w + r



7.32. For an event A,letI

equal 1 if A occurs and

let it equal 0 if A does not occur. For a random

variable X, show that

E[X|A] =

E[XI

]

P(A)

7.33. A coin that lands on heads with probability p is

continually ﬂipped. Compute the expected num-

ber of ﬂips that are made until a string of r heads

in a row is obtained.

Hint: Condition on the time of the ﬁrst occurrence

of tails to obtain the equation

E[X] = (1 − p)



i=1

i−1

(i + E[X])

+(1 − p)



i=r+1

i−1

Simplify and solve for E[X].

7.34. For another approach to Theoretical Exercise 33,

let T

denote the number of ﬂips required to obtain

a run of r consecutive heads.

(a) Determine E[T

r−1

(b) Determine E[T

]intermsofE[T

r−1

(d) What is E[T

7.35. The probability generating function of the dis-

crete nonnegative integer valued random variable

X having probability mass function p

, j Ú 0, is

deﬁned by

φ(s) = E[s

] =



j=0

Let Y be a geometric random variable with param-

eter p = 1 − s,where0 < s < 1. Suppose that Y

is independent of X, and show that

φ(s) = P{X < Y}

7.36. One ball at a time is randomly selected from an

urn containing a white and b black balls until all of

the remaining balls are of the same color. Let M

a,b

denote the expected number of balls left in the urn

when the experiment ends. Compute a recursive

formula for M

a,b

and solve when a = 3andb = 5.

7.37. An urn contains a white and b black balls. After a

ball is drawn, it is returned to the urn if it is white;

but if it is black, it is replaced by a white ball from

another urn. Let M

denote the expected number

of white balls in the urn after the foregoing opera-

tion has been repeated n times.

(a) Derive the recursive equation

n+1



1 −

a + b



+ 1

(b) Use part (a) to prove that

= a + b − b



1 −

a + b



drawn is white?

Theoretical Exercises 383

7.38. The best linear predictor of Y with respect to X

and X

is equal to a + bX

+ cX

,wherea, b,and

c are chosen to minimize

E[(Y − (a + bX

+ cX

))

]

Determine a, b,andc.

7.39. The best quadratic predictor of Y with respect to

X is a + bX + cX

,wherea, b,andc are chosen to

minimize E[(Y − (a + bX + cX

))

]. Determine

a, b,andc.

7.40. Use the conditional variance formula to determine

the variance of a geometric random variable X

having parameter p.

7.41. Let X be a normal random variable with parame-

ters μ = 0andσ

= 1, and let I, independent of X,

be such that P{I = 1}=

= P{I = 0}. Now deﬁne

Y by

Y =

X if I = 1

−X if I = 0

In words, Y is equally likely to equal either X

or −X.

(a) Are X and Y independent?

(b) Are I and Y independent?

ance 1.

(d) Show that Cov(X, Y) = 0.

7.42. It follows from Proposition 6.1 and the fact that

the best linear predictor of Y with respect to X is

+ ρ

(X − μ

) that if

E[Y|X] = a + bX

then

a = μ

− ρ

b = ρ

(Why?) Verify this directly.

7.43. Show that, for random variables X and Z,

E[(X − Y)

] = E[X

] − E[Y

]

where

Y = E [X|Z]

7.44. Consider a population consisting of individuals

able to produce offspring of the same kind. Sup-

pose that, by the end of its lifetime, each individual

will have produced j new offspring with probability

, j Ú 0, independently of the number produced

by any other individual. The number of individu-

als initially present, denoted by X

, is called the

size of the zeroth generation. All offspring of the

zeroth generation constitute the ﬁrst generation,

and their number is denoted by X

. In general,

let X

denote the size of the nth generation. Let

μ =



j=0

and σ



j=0

(j − μ)

denote, respec-

tively, the mean and the variance of the number

of offspring produced by a single individual. Sup-

pose that X

= 1—that is, initially there is a single

individual in the population.

(a) Show that

E[X

] = μE[X

n−1

]

(b) Use part (a) to conclude that

E[X

] = μ

Var(X

) = σ

n−1

+ μ

Var(X

n−1

)

(d) Use part (c) to conclude that

Var(X

) =

⎧

⎪

⎨

⎪

⎩

n−1



− 1

μ − 1



if μ Z 1

nσ

if μ = 1

The model just described is known as a

branching process, and an important ques-

tion for a population that evolves along such

lines is the probability that the population will

eventually die out. Let π denote this proba-

bility when the population starts with a single

individual. That is,

π = P{population eventually dies out|X

= 1)

(e) Argue that π satisﬁes

π =



j=0

Hint: Condition on the number of offspring of

the initial member of the population.

7.45. Verify the formula for the moment generating

function of a uniform random variable that is given

in Table 7.7. Also, differentiate to verify the for-

mulas for the mean and variance.

7.46. For a standard normal random variable Z,letμ

E[Z

]. Show that

⎧

⎪

⎨

⎪

⎩

0whenn is odd

(2j)!

when n = 2j

384 Chapter 7 Properties of Expectation

Hint: Start by expanding the moment generating

function of Z into a Taylor series about 0 to obtain

E[e

] = e



j=0

/2)

7.47. Let X be a normal random variable with mean

μ and variance σ

. Use the results of Theoretical

Exercise 46 to show that

E[X

] =

[n/2]



j=0





n−2j

(2j)!

In the preceding equation, [n/2] is the largest inte-

ger less than or equal to n/2. Check your answer by

letting n = 1andn = 2.

7.48. If Y = aX + b,wherea and b are constants,

express the moment generating function of Y in

terms of the moment generating function of X.

7.49. The positive random variable X is said to be a log-

normal random variable with parameters μ and σ

if log(X) is a normal random variable with mean

μ and variance σ

. Use the normal moment gener-

ating function to ﬁnd the mean and variance of a

lognormal random variable.

7.50. Let X have moment generating function M(t),and

deﬁne (t) = log M(t). Show that





(t)|

t=0

= Var(X)

7.51. Use Table 7.2 to determine the distribution of



i=1

when X

, ..., X

are independent and

identically distributed exponential random vari-

ables, each having mean 1/λ.

7.52. Show how to compute Cov(X, Y) from the joint

moment generating function of X and Y.

7.53. Suppose that X

, ..., X

have a multivariate nor-

mal distribution. Show that X

, ..., X

are inde-

pendent random variables if and only if

Cov(X

, X

) = 0wheni Z j

7.54. If Z is a standard normal random variable, what is

Cov(Z, Z

7.55. Suppose that Y is a normal random variable with

mean μ and variance σ

, and suppose also that the

conditional distribution of X, given that Y = y,is

normal with mean y and variance 1.

(a) Argue that the joint distribution of X, Y is the

same as that of Y + Z, Y when Z is a standard

normal random variable that is independent

of Y.

(b) Use the result of part (a) to argue that X, Y

has a bivariate normal distribution.

(d) Find E[Y|X = x].

(e) What is the conditional distribution of Y given

that X = x?

SELF-TEST PROBLEMS AND EXERCISES

7.1. Consider a list of m names, where the same name

may appear more than once on the list. Let n(i),

i = 1, ..., m, denote the number of times that the

name in position i appears on the list, and let d

denote the number of distinct names on the list.

(a) Express d in terms of the variables m, n(i), i =

1, ..., m.LetU be a uniform (0, 1) random

variable, and let X = [mU] + 1.

(b) What is the probability mass function of X?

7.2. An urn has n white and m black balls that are

removed one at a time in a randomly chosen order.

Find the expected number of instances in which a

white ball is immediately followed by a black one.

7.3. Twenty individuals consisting of 10 married cou-

ples are to be seated at 5 different tables, with 4

people at each table.

(a) If the seating is done “at random,” what is the

expected number of married couples that are

seated at the same table?

(b) If 2 men and 2 women are randomly chosen to

be seated at each table, what is the expected

number of married couples that are seated at

the same table?

7.4. If a die is to be rolled until all sides have appeared

at least once, ﬁnd the expected number of times

that outcome 1 appears.

7.5. Adeckof2n cards consists of n red and n black

cards. The cards are shufﬂed and then turned over

one at a time. Suppose that each time a red card is

turned over, we win 1 unit if more red cards than

black cards have been turned over by that time.

(For instance, if n = 2 and the result is r b r b, then

we would win a total of 2 units.) Find the expected

amount that we win.

7.6. Let A

, A

, ..., A

be events, and let N denote the

number of them that occur. Also, let I = 1ifallof

these events occur, and let it be 0 otherwise. Prove

Bonferroni’s inequality, namely,

P(A

···A

) Ú



i=1

P(A

) − (n − 1)

Hint: Argue ﬁrst that N … n − 1 + I.

Self-Test Problems and Exercises 385

7.7. Let X be the smallest value obtained when k num-

bers are randomly chosen from the set 1, ..., n.

Find E[X] by interpreting X as a negative hyper-

geometric random variable.

7.8. An arriving plane carries r families. A total of

of these families have checked in a total of j

pieces of luggage,



= r. Suppose that when

the plane lands, the N =



pieces of luggage

come out of the plane in a random order. As soon

as a family collects all of its luggage, it immediately

departs the airport. If the Sanchez family checked

in j pieces of luggage, ﬁnd the expected number of

families that depart after they do.

∗

7.9. Nineteen items on the rim of a circle of radius 1 are

to be chosen. Show that, for any choice of these

points, there will be an arc of (arc) length 1 that

contains at least 4 of them.

7.10. Let X be a Poisson random variable with mean λ.

Show that if λ is not too small, then

Var(

√

X) L .25

Hint: Use the result of Theoretical Exercise 4 to

approximate E[

√

X].

7.11. Suppose in Self-Test Problem 3 that the 20 peo-

ple are to be seated at seven tables, three of which

have 4 seats and four of which have 2 seats. If

the people are randomly seated, ﬁnd the expected

value of the number of married couples that are

seated at the same table.

7.12. Individuals 1 through n, n > 1, are to be recruited

into a ﬁrm in the following manner: Individual 1

starts the ﬁrm and recruits individual 2. Individ-

uals 1 and 2 will then compete to recruit indi-

vidual 3. Once individual 3 is recruited, individu-

als 1, 2, and 3 will compete to recruit individual 4,

and so on. Suppose that when individuals 1, 2, ..., i

compete to recruit individual i + 1, each of them

is equally likely to be the successful recruiter.

(a) Find the expected number of the individuals

1, ..., n who did not recruit anyone else.

(b) Derive an expression for the variance of the

number of individuals who did not recruit

anyone else, and evaluate it for n = 5.

7.13. The nine players on a basketball team consist of 2

centers, 3 forwards, and 4 backcourt players. If the

players are paired up at random into three groups

of size 3 each, ﬁnd (a) the expected value and (b)

the variance of the number of triplets consisting of

one of each type of player.

7.14. A deck of 52 cards is shufﬂed and a bridge hand

of 13 cards is dealt out. Let X and Y denote,

respectively, the number of aces and the number

of spades in the hand.

(a) Show that X and Y are uncorrelated.

(b) Are they independent?

7.15. Each coin in a bin has a value attached to it. Each

time that a coin with value p is ﬂipped, it lands

on heads with probability p. When a coin is ran-

domly chosen from the bin, its value is uniformly

distributed on (0, 1). Suppose that after the coin is

chosen, but before it is ﬂipped, you must predict

whether it will land on heads or on tails. You will

win 1 if you are correct and will lose 1 otherwise.

(a) What is your expected gain if you are not told

the value of the coin?

(b) Suppose now that you are allowed to inspect

the coin before it is ﬂipped, with the result of

your inspection being that you learn the value

of the coin. As a function of p, the value of the

coin, what prediction should you make?

expected gain?

7.16. In Self-Test Problem 1, we showed how to use

the value of a uniform (0, 1) random variable

(commonly called a random number) to obtain the

value of a random variable whose mean is equal

to the expected number of distinct names on a list.

However, its use required that one choose a ran-

dom position and then determine the number of

times that the name in that position appears on

the list. Another approach, which can be more efﬁ-

cient when there is a large amount of replication of

names, is as follows: As before, start by choosing

the random variable X

as in Problem 1. Now iden-

tify the name in position X, and then go through

the list, starting at the beginning, until that name

appears. Let I equal 0 if you encounter that name

before getting to position X,andletI equal 1 if

your ﬁrst encounter with the name is at position X.

Show that E[mI] = d.

Hint: Compute E[I] by using conditional expecta-

tion.

7.17. A total of m items are to be sequentially dis-

tributed among n cells, with each item indepen-

dently being put in cell j with probability p

, j =

1, ..., n. Find the expected number of collisions

that occur, where a collision occurs whenever an

item is put into a nonempty cell.

7.18. Let X be the length of the initial run in a random

ordering of n ones and m zeroes. That is, if the ﬁrst

k values are the same (either all ones or all zeroes),

then X Ú k.FindE[X].

7.19. There are n items in a box labeled H and m in a

box labeled T. A coin that comes up heads with

probability p and tails with probability 1 − p is

ﬂipped. Each time it comes up heads, an item is

removed from the H box, and each time it comes

up tails, an item is removed from the T box. (If a

box is empty and its outcome occurs, then no items

386 Chapter 7 Properties of Expectation

are removed.) Find the expected number of coin

ﬂips needed for both boxes to become empty.

Hint: Condition on the number of heads in the ﬁrst

n + m ﬂips.

7.20. Let X be a nonnegative random variable having

distribution function F. Show that if

F(x) = 1 −

F(x),then

E[X

] =

n−1

F(x) dx

Hint: Start with the identity

= n

n−1

= n

n−1

(x) dx

where

(x) =

1, if x < X

0, otherwise

∗

7.21. Let a

, ..., a

, not all equal to 0, be such that



i=1

= 0. Show that there is a permutation

, ..., i

such that



j=1

j+1

< 0.

Hint: Use the probabilistic method. (It is interest-

ing that there need not be a permutation whose

sum of products of successive pairs is positive. For

instance, if n = 3, a

= a

=−1, and a

= 2, there

is no such permutation.)

7.22. Suppose that X

, i = 1, 2, 3, are independent Pois-

son random variables with respective means λ

i = 1, 2, 3. Let X = X

+ X

and Y = X

+ X

The random vector X, Y is said to have a bivariate

Poisson distribution.

(a) Find E[X]andE[Y].

(b) Find Cov(X, Y).

P{X = i, Y = j}.

7.23. Let (X

, Y

), i = 1, ..., be a sequence of indepen-

dent and identically distributed random vectors.

That is, X

, Y

is independent of, and has the same

distribution as X

, Y

, and so on. Although X

and

can be dependent, X

and Y

are independent

when i Z j.Let

= E[X

], μ

= E[Y

], σ

= Var(X

= Var(Y

), ρ = Corr(X

, Y

)

Find Corr(



i=1



j=1

7.24. Three cards are randomly chosen without replace-

ment from an ordinary deck of 52 cards. Let X

denote the number of aces chosen.

(a) Find E[X|the ace of spades is chosen].

(b) Find E[X|at least one ace is chosen].

7.25. Let  be the standard normal distribution func-

tion, and let X be a normal random variable with

mean μ and variance 1. We want to ﬁnd E[(X)].

To do so, let Z be a standard normal random vari-

able that is independent of X,andlet

I =

1, if Z < X

0, if Z Ú X

(a) Show that E[I|X = x] = (x).

(b) Show that E[(X)] = P{Z < X}.

√

Hint: What is the distribution of X − Z?

The preceding comes up in statistics. Suppose

you are about to observe the value of a random

variable X that is normally distributed with an

unknown mean μ and variance 1, and suppose that

you want to test the hypothesis that the mean μ

is greater than or equal to 0. Clearly you would

want to reject this hypothesis if X is sufﬁciently

small. If it results that X = x, then the p-value

of the hypothesis that the mean is greater than

or equal to 0 is deﬁned to be the probability that

X would be as small as x if μ were equal to 0

(its smallest possible value if the hypothesis were

true). (A small p-value is taken as an indication

that the hypothesis is probably false.) Because X

has a standard normal distribution when μ = 0,

the p-value that results when X = x is (x).

Therefore, the preceding shows that the expected

p-value that results when the true mean is μ

is (

√

7.26. A coin that comes up heads with probability p

is ﬂipped until either a total of n heads or of

m tails is amassed. Find the expected number of

ﬂips.

Hint: Imagine that one continues to ﬂip even after

the goal is attained. Let X denote the number of

ﬂips needed to obtain n heads, and let Y denote

the number of ﬂips needed to obtain m tails. Note

that max(X, Y) + min(X, Y) = X + Y. Com-

pute E[max(X, Y)] by conditioning on the number

of heads in the ﬁrst n + m − 1ﬂips.

7.27. Adeckofn cards numbered 1 through n, initially

in any arbitrary order, is shufﬂed in the following

manner: At each stage, we randomly choose one

of the cards and move it to the front of the deck,

leaving the relative positions of the other cards

unchanged. This procedure is continued until all

but one of the cards has been chosen. At this point

it follows by symmetry that all n! possible order-

ings are equally likely. Find the expected number

of stages that are required.

7.28. Suppose that a sequence of independent trials in

which each trial is a success with probability p is

Self-Test Problems and Exercises 387

performed until either a success occurs or a total of

n trials has been reached. Find the mean number

of trials that are performed.

Hint: The computations are simpliﬁed if you use

the identity that, for a nonnegative integer valued

random variable X,

E[X] =



i=1

P{X Ú i}

7.29. Suppose that X and Y are both Bernoulli random

variables. Show that X and Y are independent if

and only if Cov(X, Y) = 0.

7.30. In the generalized match problem, there are n

individuals of whom n

wear hat size i,



i=1

n. There are also n hats, of which h

are of

size i,



i=1

= n. If each individual randomly

chooses a hat (without replacement), ﬁnd the

expected number who choose a hat that is their

size.

CHAPTER 8

Limit Theorems

8.1 INTRODUCTION

8.2 CHEBYSHEV’S INEQUALITY AND THE WEAK LAW OF LARGE NUMBERS

8.3 THE CENTRAL LIMIT THEOREM

8.4 THE STRONG LAW OF LARGE NUMBERS

8.5 OTHER INEQUALITIES

8.6 BOUNDING THE ERROR PROBABILITY WHEN APPROXIMATING A SUM OF INDEPENDENT BERNOULLI

RANDOM VARIABLES BY A POISSON RANDOM VARIABLE

8.1 INTRODUCTION

The most important theoretical results in probability theory are limit theorems. Of

these, the most important are those classiﬁed either under the heading laws of large

numbers or under the heading central limit theorems. Usually, theorems are consid-

ered to be laws of large numbers if they are concerned with stating conditions under

which the average of a sequence of random variables converges (in some sense) to

the expected average. By contrast, central limit theorems are concerned with deter-

mining conditions under which the sum of a large number of random variables has a

probability distribution that is approximately normal.

8.2 CHEBYSHEV’S INEQUALITY AND THE WEAK LAW OF LARGE NUMBERS

We start this section by proving a result known as Markov’s inequality.

Proposition 2.1. Markov’s inequality

If X is a random variable that takes only nonnegative values, then, for any value

a > 0,

P{X Ú a} …

E[X]

Proof. For a > 0, let

I =

1ifX Ú a

0 otherwise

and note that, since X Ú 0,

I …

Taking expectations of the preceding inequality yields

E[I] …

E[X]

which, because E[I] = P{X Ú a}, proves the result.

As a corollary, we obtain Proposition 2.2.

388

Section 8.2 Chebyshev’s Inequality and the Weak Law of Large Numbers 389

Proposition 2.2. Chebyshev’s inequality

If X is a random variable with ﬁnite mean μ and variance σ

, then, for any value

k > 0,

P{|X − μ| Ú k} …

Proof. Since (X − μ)

is a nonnegative random variable, we can apply Markov’s

inequality (with a = k

) to obtain

P{(X − μ)

Ú k

} …

E[(X − μ)

]

(2.1)

But since (X − μ)

Ú k

if and only if |X − μ| Ú k, Equation (2.1) is equivalent to

P{|X − μ| Ú k} …

E[(X − μ)

]

and the proof is complete.

The importance of Markov’s and Chebyshev’s inequalities is that they enable us to

derive bounds on probabilities when only the mean, or both the mean and the vari-

ance, of the probability distribution are known. Of course, if the actual distribution

were known, then the desired probabilities could be computed exactly and we would

not need to resort to bounds.

EXAMPLE 2a

Suppose that it is known that the number of items produced in a factory during a

week is a random variable with mean 50.

(a) What can be said about the probability that this week’s production will

exceed 75?

(b) If the variance of a week’s production is known to equal 25, then what can

be said about the probability that this week’s production will be between 40

and 60?

Solution. Let X be the number of items that will be produced in a week.

(a) By Markov’s inequality,

P{X > 75} …

E[X]

(b) By Chebyshev’s inequality,

P{|X − 50| Ú 10} …

Hence,

P{|X − 50| < 10} Ú 1 −

so the probability that this week’s production will be between 40 and 60 is at

least .75. .

390 Chapter 8 Limit Theorems

As Chebyshev’s inequality is valid for all distributions of the random variable X,

we cannot expect the bound on the probability to be very close to the actual proba-

bility in most cases. For instance, consider Example 2b.

EXAMPLE 2b

If X is uniformly distributed over the interval (0, 10), then, since E[X] = 5and

Var(X) =

, it follows from Chebyshev’s inequality that

P{|X − 5| > 4} …

3(16)

L .52

whereas the exact result is

P{|X − 5| > 4}=.20

Thus, although Chebyshev’s inequality is correct, the upper bound that it provides is

not particularly close to the actual probability.

Similarly, if X is a normal random variable with mean μ and variance σ

Chebyshev’s inequality states that

P{|X − μ| > 2σ } …

whereas the actual probability is given by

P{|X − μ| > 2σ }=P

X − μ

> 2

= 2[1 − (2)] L .0456 .

Chebyshev’s inequality is often used as a theoretical tool in proving results. This

use is illustrated ﬁrst by Proposition 2.3 and then, most importantly, by the weak law

of large numbers.

Proposition 2.3. If Var(X) = 0, then

P{X = E[X]}=1

In other words, the only random variables having variances equal to 0 are those which

are constant with probability 1.

Proof. By Chebyshev’s inequality, we have, for any n Ú 1,

|X − μ| >

= 0

Letting n

→q and using the continuity property of probability yields

0 = lim

n→q

|X − μ| >

= P

lim

n→q

|X − μ| >

= P{X Z μ}

and the result is established.

Section 8.3 The Central Limit Theorem 391

Theorem 2.1 The weak law of large numbers

Let X

, X

, ...be a sequence of independent and identically distributed random vari-

ables, each having ﬁnite mean E[X

] = μ. Then, for any ε>0,

+ ··· + X

− μ

Ú ε

→0 as n→q

Proof. We shall prove the theorem only under the additional assumption that the

random variables have a ﬁnite variance σ

. Now, since



+ ··· + X



= μ and Var



+ ··· + X



it follows from Chebyshev’s inequality that

+ ··· + X

− μ

Ú ε

…

nε

and the result is proven.

The weak law of large numbers was originally proven by James Bernoulli for the

special case where the X

are 0, 1 (that is, Bernoulli) random variables. His statement

and proof of this theorem were presented in his book Ars Conjectandi, which was

published in 1713, eight years after his death, by his nephew Nicholas Bernoulli. Note

that, because Chebyshev’s inequality was not known in Bernoulli’s time, Bernoulli

had to resort to a quite ingenious proof to establish the result. The general form of

the weak law of large numbers presented in Theorem 2.1 was proved by the Russian

mathematician Khintchine.

8.3 THE CENTRAL LIMIT THEOREM

The central limit theorem is one of the most remarkable results in probability theory.

Loosely put, it states that the sum of a large number of independent random variables

has a distribution that is approximately normal. Hence, it not only provides a simple

method for computing approximate probabilities for sums of independent random

variables, but also helps explain the remarkable fact that the empirical frequencies of

so many natural populations exhibit bell-shaped (that is, normal) curves.

In its simplest form the central limit theorem is as follows.

Theorem 3.1 The central limit theorem

Let X

, X

, ...be a sequence of independent and identically distributed random vari-

ables, each having mean μ and variance σ

. Then the distribution of

+ ··· + X

− nμ

√

tends to the standard normal as n

→q. That is, for −q < a < q,

+ ··· + X

− nμ

√

… a

→

√

2π

−q

−x

dx as n→q

The key to the proof of the central limit theorem is the following lemma, which we

state without proof.

392 Chapter 8 Limit Theorems

Lemma 3.1

Let Z

, Z

, ...be a sequence of random variables having distribution functions F

and moment generating functions M

, n Ú 1; and let Z be a random variable

having distribution function F

and moment generating function M

.IfM

(t) →

(t) for all t,thenF

(t) →F

(t) for all t at which F

(t) is continuous.

If we let Z be a standard normal random variable, then, since M

(t) = e

,it

follows from Lemma 3.1 that if M

(t) →e

as n →q,thenF

(t) →(t) as n →q.

We are now ready to prove the central limit theorem.

Proof of the Central Limit Theorem: Let us assume at ﬁrst that μ = 0andσ

= 1.

We shall prove the theorem under the assumption that the moment generating func-

tion of the X

, M(t), exists and is ﬁnite. Now, the moment generating function of

√

n is given by

⎡

⎣

exp

√

⎤

⎦

= M



√



Thus, the moment generating function of



i=1

√

n is given by





√





.Let

L(t) = log M(t)

and note that

L(0) = 0



(0) =



(0)

M(0)

= μ

= 0



(0) =

M(0)M



(0) − [M



(0)]

[M(0)]

= E[X

]

= 1

Now, to prove the theorem, we must show that [M(t/

√

n)]

→e

as n →q,or,

equivalently, that nL(t/

√

n) →t

/2asn →q. To show this, note that

lim

n→q

L(t/

√

−1

= lim

n→q

−L



(t/

√

n)n

−3/2

−2n

−2

by L’H

opital’s rule

= lim

n→q





(t/

√

n)t

−1/2



= lim

n→q



−L



(t/

√

n)n

−3/2

−2n

−3/2



again by L’H

opital’s rule

= lim

n→q

⎡

⎣





√



⎤

⎦

Section 8.3 The Central Limit Theorem 393

Thus, the central limit theorem is proven when μ = 0andσ

= 1. The result

now follows in the general case by considering the standardized random variables

∗

= (X

− μ)/σ and applying the preceding result, since E[X

∗

] = 0, Var(X

∗

) = 1.

Remark. Although Theorem 3.1 states only that, for each a,

+ ··· + X

− nμ

√

… a

→(a)

it can, in fact, be shown that the convergence is uniform in a. [We say that f

(a) →f (a)

uniformly in a if, for each ε>0, there exists an N such that |f

(a) − f (a)| <εfor all

a whenever n Ú N.] .

The ﬁrst version of the central limit theorem was proven by DeMoivre around

1733 for the special case where the X

are Bernoulli random variables with p =

The theorem was subsequently extended by Laplace to the case of arbitrary p. (Since

a binomial random variable may be regarded as the sum of n independent and identi-

cally distributed Bernoulli random variables, this justiﬁes the normal approximation

to the binomial that was presented in Section 5.4.1.) Laplace also discovered the more

general form of the central limit theorem given in Theorem 3.1. His proof, however,

was not completely rigorous and, in fact, cannot easily be made rigorous. A truly

rigorous proof of the central limit theorem was ﬁrst presented by the Russian mathe-

matician Liapounoff in the period 1901–1902.

This important theorem is illustrated by the central limit theorem module on the

text website. This website yields plots of the density function of the sum of n inde-

pendent and identically distributed random variables that each take on one of the

values 0, 1, 2, 3, 4. When using it, one enters the probability mass function and the

desired value of n. Figure 8.1 shows the resulting plots for a speciﬁed probability mass

function when (a) n = 5, (b) n = 10, (c) n = 25, and (d) n = 100.

EXAMPLE 3a

An astronomer is interested in measuring the distance, in light-years, from his obser-

vatory to a distant star. Although the astronomer has a measuring technique, he

knows that, because of changing atmospheric conditions and normal error, each time

a measurement is made it will not yield the exact distance, but merely an estimate.

As a result, the astronomer plans to make a series of measurements and then use the

average value of these measurements as his estimated value of the actual distance.

If the astronomer believes that the values of the measurements are independent and

identically distributed random variables having a common mean d (the actual dis-

tance) and a common variance of 4 (light-years), how many measurements need he

make to be reasonably sure that his estimated distance is accurate to within ;.5 light-

year?

Solution. Suppose that the astronomer decides to make n observations. If X

, ..., X

are the n measurements, then, from the central limit theorem, it

follows that



i=1

− nd

√

394 Chapter 8 Limit Theorems

Central Limit Theorem

Enter the probabilities and the number of random

variables to be summed. The output gives the mass

function of the sum along with its mean and

variance.

P0 .25

P1 .15

P2 .1

P3 .2

P4 .3

n = 5

Start

Quit

Mean = 10.75

Variance = 12.6375

20151050

0.00

0.05

0.10

0.15

p(i)

FIGURE 8.1(a)

has approximately a standard normal distribution. Hence,

⎧

⎪

⎨

⎪

⎩

− .5 …



i=1

− d … .5

⎫

⎪

⎬

⎪

⎭

= P

−.5

√

… Z

… .5

√

L 



√



− φ



−

√



= 2



√



− 1

Therefore, if the astronomer wants, for instance, to be 95 percent certain that his

estimated value is accurate to within .5 light year, he should make n

∗

measurements,

where n

∗

is such that

2



√

∗



− 1 = .95 or 



√

∗



= .975

Thus, from Table 5.1 of Chapter 5,

√

∗

= 1.96 or n

∗

= (7.84)

L 61.47

As n

∗

is not integral valued, he should make 62 observations.

Section 8.3 The Central Limit Theorem 395

Central Limit Theorem

Enter the probabilities and the number of random

variables to be summed. The output gives the mass

function of the sum along with its mean and

variance.

P0 .25

P1 .15

P2 .1

P3 .2

P4 .3

n = 10

Start

Quit

Mean = 21.5

Variance = 25.275

403020100

0.00

0.08

p(i)

0.06

0.04

0.02

FIGURE 8.1(b)

Note, however, that the preceding analysis has been done under the assumption

that the normal approximation will be a good approximation when n = 62. Although

this will usually be the case, in general the question of how large n need be before

the approximation is “good” depends on the distribution of the X

. If the astronomer

is concerned about this point and wants to take no chances, he can still solve his

problem by using Chebyshev’s inequality. Since

⎡

⎣



i=1

⎤

⎦

= d Var

⎛

⎝



i=1

⎞

⎠

Chebyshev’s inequality yields

⎧

⎪

⎨

⎪

⎩



i=1

− d

>.5

⎫

⎪

⎬

⎪

⎭

…

n(.5)

Hence, if he makes n = 16/.05 = 320 observations, he can be 95 percent certain that

his estimate will be accurate to within .5 light-year. .

396 Chapter 8 Limit Theorems

Central Limit Theorem

Enter the probabilities and the number of random

variables to be summed. The output gives the mass

function of the sum along with its mean and

variance.

P0 .25

P1 .15

P2 .1

P3 .2

P4 .3

n = 25

Start

Quit

Mean = 53.75

Variance = 63.1875

1000

0.00

0.05

p(i)

0.04

0.03

0.02

0.01

80604020

FIGURE 8.1(c)

EXAMPLE 3b

The number of students who enroll in a psychology course is a Poisson random vari-

able with mean 100. The professor in charge of the course has decided that if the

number enrolling is 120 or more, he will teach the course in two separate sections,

whereas if fewer than 120 students enroll, he will teach all of the students together

in a single section. What is the probability that the professor will have to teach two

sections?

Solution. The exact solution

−100



i=120

(100)

does not readily yield a numerical answer. However, by recalling that a Poisson ran-

dom variable with mean 100 is the sum of 100 independent Poisson random variables,

each with mean 1, we can make use of the central limit theorem to obtain an approx-

imate solution. If X denotes the number of students that enroll in the course, we

have

P{X Ú 120}=P{X Ú 119.5} (the continuity correction)

Section 8.3 The Central Limit Theorem 397

Central Limit Theorem

Enter the probabilities and the number of random

variables to be summed. The output gives the mass

function of the sum along with its mean and

variance.

P0 .25

P1 .15

P2 .1

P3 .2

P4 .3

n = 100

Start

Quit

Mean = 215.

Variance = 252.75

4000

0.000

0.030

p(i)

300200100

0.025

0.020

0.015

0.010

0.005

FIGURE 8.1(d)

= P

X − 100

√

100

119.5 − 100

√

100

L 1 − (1.95)

L .0256

where we have used the fact that the variance of a Poisson random variable is equal

to its mean. .

EXAMPLE 3c

If 10 fair dice are rolled, ﬁnd the approximate probability that the sum obtained is

between 30 and 40, inclusive.

Solution. Let X

denote the value of the ith die, i = 1, 2, ..., 10. Since

E(X

) =

,Var(X

) = E[X

] − (E[X

])

398 Chapter 8 Limit Theorems

the central limit theorem yields

P{29.5 … X … 40.5}=P

⎧

⎪

⎨

⎪

⎩

29.5 − 35

350

…

X − 35

350

…

40.5 − 35

350

⎫

⎪

⎬

⎪

⎭

L 2(1.0184) − 1

L .692 .

EXAMPLE 3d

Let X

, i = 1, ..., 10, be independent random variables, each uniformly distributed

over (0, 1). Calculate an approximation to P



i=1

> 6

Solution. Since E[X

] =

and Var(X

) =

, we have, by the central limit theorem,

⎧

⎨

⎩



> 6

⎫

⎬

⎭

= P

⎧

⎪

⎨

⎪

⎩



− 5

10(

)

6 − 5

10(

)

⎫

⎪

⎬

⎪

⎭

L 1 − (

√

1.2)

L .1367

Hence,



i=1

will be greater than 6 only 14 percent of the time. .

EXAMPLE 3e

An instructor has 50 exams that will be graded in sequence. The times required to

grade the 50 exams are independent, with a common distribution that has mean

20 minutes and standard deviation 4 minutes. Approximate the probability that the

instructor will grade at least 25 of the exams in the ﬁrst 450 minutes of work.

Solution. If we let X

be the time that it takes to grade exam i,then

X =



i=1

is the time it takes to grade the ﬁrst 25 exams. Because the instructor will grade at

least 25 exams in the ﬁrst 450 minutes of work if the time it takes to grade the ﬁrst 25

exams is less than or equal to 450, we see that the desired probability is P{X … 450}.

To approximate this probability, we use the central limit theorem. Now,

E[X] =



i=1

E[X

] = 25(20) = 500

and

Var(X) =



i=1

Var(X

) = 25(16) = 400

Section 8.3 The Central Limit Theorem 399

Consequently, with Z being a standard normal random variable, we have

P{X … 450}=P{

X − 500

√

400

…

450 − 500

√

400

}

L P{Z … −2.5}

= P{Z Ú 2.5}

= 1 − (2.5) = .006

Central limit theorems also exist when the X

are independent, but not necessarily

identically distributed random variables. One version, by no means the most general,

is as follows.

Theorem 3.2 Central limit theorem for independent random variables

Let X

, X

, ... be a sequence of independent random variables having respective

means and variances μ

= E[X

], σ

= Var(X

). If (a) the X

are uniformly

bounded—that is, if for some M, P{|X

| < M}=1 for all i, and (b)



i=1

=q—then

⎧

⎪

⎨

⎪

⎩



i=1

− μ

)



i=1

… a

⎫

⎪

⎬

⎪

⎭

→(a) as n →q

Historical Note

Pierre-Simon, Marquis de Laplace

The central limit theorem was originally stated and proven by the French math-

ematician Pierre-Simon, Marquis de Laplace, who came to the theorem from his

observations that errors of measurement (which can usually be regarded as being

the sum of a large number of tiny forces) tend to be normally distributed. Laplace,

who was also a famous astronomer (and indeed was called “the Newton of France”),

was one of the great early contributors to both probability and statistics. Laplace

was also a popularizer of the uses of probability in everyday life. He strongly believed

in its importance, as is indicated by the following quotations of his taken from his

published book Analytical Theory of Probability: “We see that the theory of proba-

bility is at bottom only common sense reduced to calculation; it makes us appreciate

with exactitude what reasonable minds feel by a sort of instinct, often without being

able to account for it.... It is remarkable that this science, which originated in the

consideration of games of chance, should become the most important object of

human knowledge.... The most important questions of life are, for the most part,

really only problems of probability.”

The application of the central limit theorem to show that measurement errors

are approximately normally distributed is regarded as an important contribution to

science. Indeed, in the 17th and 18th centuries the central limit theorem was often

called the law of frequency of errors. Listen to the words of Francis Galton (taken

from his book Natural Inheritance, published in 1889): “I know of scarcely anything

so apt to impress the imagination as the wonderful form of cosmic order expressed

by the ‘Law of Frequency of Error.’ The Law would have been personiﬁed by the

Greeks and deiﬁed, if they had known of it. It reigns with serenity and in complete

self-effacement amidst the wildest confusion. The huger the mob and the greater the

apparent anarchy, the more perfect is its sway. It is the supreme law of unreason.”

400 Chapter 8 Limit Theorems

8.4 THE STRONG LAW OF LARGE NUMBERS

The strong law of large numbers is probably the best-known result in probability the-

ory. It states that the average of a sequence of independent random variables having

a common distribution will, with probability 1, converge to the mean of that distribu-

tion.

Theorem 4.1 The strong law of large numbers

Let X

, X

, ...be a sequence of independent and identically distributed random vari-

ables, each having a ﬁnite mean μ = E[X

]. Then, with probability 1,

+ X

+ ··· + X

→μ as n →q

†

As an application of the strong law of large numbers, suppose that a sequence of

independent trials of some experiment is performed. Let E be a ﬁxed event of the

experiment, and denote by P(E) the probability that E occurs on any particular trial.

Letting

1ifE occurs on the ith trial

0ifE does not occur on the ith trial

we have, by the strong law of large numbers, that with probability 1,

+ ··· + X

→E[X] = P(E) (4.1)

Since X

+ ··· + X

represents the number of times that the event E occurs in the

ﬁrst n trials, we may interpret Equation (4.1) as stating that, with probability 1, the

limiting proportion of time that the event E occurs is just P(E).

Although the theorem can be proven without this assumption, our proof of the

strong law of large numbers will assume that the random variables X

have a ﬁnite

fourth moment. That is, we will suppose that E[X

] = K < q.

Proof of the Strong Law of Large Numbers: To begin, assume that μ, the mean of

the X

, is equal to 0. Let S



i=1

and consider

E[S

] = E[(X

+ ··· + X

)(X

+ ··· + X

)

* (X

+ ··· + X

)(X

+ ··· + X

)]

Expanding the right side of the preceding equation results in terms of the form

, X

,andX

where i, j, k,andl are all different. Because all the X

have mean 0, it follows by

independence that

E[X

] = E[X

]E[X

] = 0

E[X

] = E[X

]E[X

] = 0

E[X

] = 0

†

That is, the strong law of large numbers states that

P{ lim

n→q

+ ··· + X

)/n = μ}=1

Section 8.4 The Strong Law of Large Numbers 401

Now, for a given pair i and j, there will be





= 6 terms in the expansion that will

equal X

. Hence, upon expanding the preceding product and taking expectations

term by term, it follows that

E[S

] = nE[X

] + 6





E[X

]

= nK + 3n(n − 1)E[X

]E[X

]

where we have once again made use of the independence assumption. Now, since

0 … Var(X

) = E[X

] − (E[X

])

we have

(E[X

])

… E[X

] = K

Therefore, from the preceding, we obtain

E[S

] … nK + 3n(n − 1)K

which implies that





…

Therefore,

⎡

⎣



n=1

⎤

⎦



n=1





< q

But the preceding implies that, with probability 1,



n=1

< q. (For if there is a

positive probability that the sum is inﬁnite, then its expected value is inﬁnite.) But the

convergence of a series implies that its nth term goes to 0; so we can conclude that,

with probability 1,

lim

n→q

= 0

But if S

= (S

/n)

goes to 0, then so must S

/n; hence, we have proven that, with

probability 1,

→0asn →q

When μ, the mean of the X

, is not equal to 0, we can apply the preceding argument

to the random variables X

− μ to obtain that with probability 1,

lim

n→q



i=1

− μ)

= 0

or, equivalently,

lim

n→q



i=1

= μ

which proves the result.

402 Chapter 8 Limit Theorems

The strong law is illustrated by two modules on the text website that consider inde-

pendent and identically distributed random variables which take on one of the values

0, 1, 2, 3, and 4. The modules simulate the values of n such random variables; the

proportions of time that each outcome occurs, as well as the resulting sample mean



i=1

/n, are then indicated and plotted. When using these modules, which differ only

in the type of graph presented, one enters the probabilities and the desired value of n.

Figure 8.2 gives the results of a simulation using a speciﬁed probability mass function

and (a) n = 100, (b) n = 1000, and (c) n = 10,000.

Many students are initially confused about the difference between the weak and

the strong laws of large numbers. The weak law of large numbers states that, for any

speciﬁed large value n

∗

, (X

+ ··· + X

∗

)/n

∗

is likely to be near μ. However, it does

not say that (X

+ ··· + X

)/n is bound to stay near μ for all values of n larger than

∗

. Thus, it leaves open the possibility that large values of |(X

+ ··· + X

)/n − μ|

can occur inﬁnitely often (though at infrequent intervals). The strong law shows that

this cannot occur. In particular, it implies that, with probability 1, for any positive

value ε,



− μ

will be greater than ε only a ﬁnite number of times.

Strong Law Of Large Numbers

Enter the probabilities and the number of trials

to be simulated. The output gives the total number

of times each outcome occurs, and the average

of all outcomes.

P0 .1

P1 .2

P2 .3

P3 .35

P4 .05

n = 100

Start

Quit

Theoretical Mean = 2.05

Sample Mean = 1.89

FIGURE 8.2(a)

Section 8.5 Other Inequalities 403

Strong Law Of Large Numbers

Enter the probabilities and the number of trials

to be simulated. The output gives the total number

of times each outcome occurs, and the average

of all outcomes.

P0 .1

P1 .2

P2 .3

P3 .35

P4 .05

n = 1000

Start

Quit

Theoretical Mean = 2.05

Sample Mean = 2.078

106

189

285

361

FIGURE 8.2(b)

The strong law of large numbers was originally proven, in the special case of

Bernoulli random variables, by the French mathematician Borel. The general form of

the strong law presented in Theorem 4.1 was proven by the Russian mathematician

A. N. Kolmogorov.

8.5 OTHER INEQUALITIES

We are sometimes confronted with situations in which we are interested in obtaining

an upper bound for a probability of the form P{X − μ Ú a}, where a is some positive

value and when only the mean μ = E[X] and variance σ

= Var(X) of the distribu-

tion of X are known. Of course, since X − μ Ú a > 0 implies that |X − μ| Ú a,it

follows from Chebyshev’s inequality that

P{X − μ Ú a} … P{|X − μ| Ú a} …

when a > 0

However, as the following proposition shows, it turns out that we can do better.

Proposition 5.1. One-sided Chebyshev inequality

If X is a random variable with mean 0 and ﬁnite variance σ

, then, for any a > 0,

P{X Ú a} …

+ a

404 Chapter 8 Limit Theorems

Strong Law Of Large Numbers

Enter the probabilities and the number of trials

to be simulated. The output gives the total number

of times each outcome occurs, and the average

of all outcomes.

P0 .1

P1 .2

P2 .3

P3 .35

P4 .05

n = 10000

Start

Quit

Theoretical Mean = 2.05

Sample Mean = 2.0416

1041

2027

2917

3505

510

FIGURE 8.2(c)

Proof. Let b > 0 and note that

X Ú a is equivalent to X + b Ú a + b

Hence,

P{X Ú a}=P{X + b Ú a + b}

… P{(X + b)

Ú (a + b)

}

where the inequality is obtained by noting that since a + b > 0, X + b Ú a + b

implies that (X + b)

Ú (a + b)

. Upon applying Markov’s inequality, the

preceding yields that

P{X Ú a} …

E[(X + b)

]

(a + b)

+ b

(a + b)

Letting b = σ

/a [which is easily seen to be the value of b that minimizes

(σ

+ b

)/(a + b)

] gives the desired result.

Section 8.5 Other Inequalities 405

EXAMPLE 5a

If the number of items produced in a factory during a week is a random variable with

mean 100 and variance 400, compute an upper bound on the probability that this

week’s production will be at least 120.

Solution. It follows from the one-sided Chebyshev inequality that

P{X Ú 120}=P{X − 100 Ú 20} …

400

400 + (20)

Hence, the probability that this week’s production will be 120 or more is at most

If we attempted to obtain a bound by applying Markov’s inequality, then we would

have obtained

P{X Ú 120} …

E(X)

120

which is a far weaker bound than the preceding one. .

Suppose now that X has mean μ and variance σ

. Since both X − μ and μ − X

have mean 0 and variance σ

, it follows from the one-sided Chebyshev inequality

that, for a > 0,

P{X − μ Ú a} …

+ a

and

P{μ − X Ú a} …

+ a

Thus, we have the following corollary.

Corollary 5.1. If E[X] = μ and Var(X) = σ

, then, for a > 0,

P{X Ú μ + a} …

+ a

P{X … μ − a} …

+ a

EXAMPLE 5b

A set of 200 people consisting of 100 men and 100 women is randomly divided into

100 pairs of 2 each. Give an upper bound to the probability that at most 30 of these

pairs will consist of a man and a woman.

Solution. Number the men arbitrarily from 1 to 100, and for i = 1, 2, ...100, let

1ifmani is paired with a woman

0 otherwise

Then X, the number of man–woman pairs, can be expressed as

X =

100



i=1

406 Chapter 8 Limit Theorems

Because man i is equally likely to be paired with any of the other 199 people, of which

100 are women, we have

E[X

] = P{X

= 1}=

100

199

Similarly, for i Z j,

E[X

] = P{X

= 1, X

= 1}

= P{X

= 1}P{X

= 1|X

= 1}=

100

199

197

where P{X

= 1|X

= 1}=99/197, since, given that man i is paired with a woman,

man j is equally likely to be paired with any of the remaining 197 people, of which 99

are women. Hence, we obtain

E[X] =

100



i=1

E[X

]

= (100)

100

199

L 50.25

Var(X) =

100



i=1

Var(X

) + 2



i<j



Cov(X

, X

)

= 100

100

199

+ 2



100





100

199

197

−



100

199





L 25.126

The Chebyshev inequality then yields

P{X … 30} … P{|X − 50.25| Ú 20.25} …

25.126

(20.25)

L .061

Thus, there are fewer than 6 chances in a hundred that fewer than 30 men will be

paired with women. However, we can improve on this bound by using the one-sided

Chebyshev inequality, which yields

P{X … 30}=P{X … 50.25 − 20.25}

…

25.126

25.126 + (20.25)

L .058 .

When the moment generating function of the random variable X is known, we can

obtain even more effective bounds on P{X Ú a}.Let

M(t) = E[e

]

be the moment generating function of the random variable X. Then, for t > 0,

P{X Ú a}=P{e

Ú e

}

… E[e

−ta

by Markov’s inequality

Section 8.5 Other Inequalities 407

Similarly, for t < 0,

P{X … a}=P{e

Ú e

}

… E[e

−ta

Thus, we have the following inequalities, known as Chernoff bounds.

Proposition 5.2. Chernoff bounds

P{X Ú a} … e

−ta

M(t) for all t > 0

P{X … a} … e

−ta

M(t) for all t < 0

Since the Chernoff bounds hold for all t in either the positive or negative quadrant,

we obtain the best bound on P{X Ú a} by using the t that minimizes e

−ta

M(t).

EXAMPLE 5c Chernoff bounds for the standard normal random variable

If Z is a standard normal random variable, then its moment generating function is

M(t) = e

, so the Chernoff bound on P{Z Ú a} is given by

P{Z Ú a} … e

−ta

for all t > 0

Now the value of t, t > 0, that minimizes e

/2−ta

is the value that minimizes t

/2 − ta,

which is t = a. Thus, for a > 0, we have

P{Z Ú a} … e

−a

Similarly, we can show that, for a < 0,

P{Z … a} … e

−a

EXAMPLE 5d Chernoff bounds for the Poisson random variable

If X is a Poisson random variable with parameter λ, then its moment generating func-

tion is M(t) = e

λ(e

−1)

. Hence, the Chernoff bound on P{X Ú i} is

P{X Ú i} … e

λ(e

−1)

−it

t > 0

Minimizing the right side of the preceding inequality is equivalent to minimizing

λ(e

− 1) − it, and calculus shows that the minimal value occurs when e

= i/λ. Pro-

vided that i/λ > 1, this minimizing value of t will be positive. Therefore, assuming

that i >λand letting e

= i/λ in the Chernoff bound yields

P{X Ú i} … e

λ(i/λ−1)





or, equivalently,

P{X Ú i} …

−λ

(eλ)

EXAMPLE 5e

Consider a gambler who is equally likely to either win or lose 1 unit on every play,

independently of his past results. That is, if X

is the gambler’s winnings on the ith

play, then the X

are independent and

408 Chapter 8 Limit Theorems

P{X

= 1}=P{X

=−1}=

Let S



i=1

denote the gambler’s winnings after n plays. We will use the Chernoff

bound on P{S

Ú a}. To start, note that the moment generating function of X

E[e

] =

+ e

−t

Now, using the McLaurin expansions of e

and e

−t

, we see that

+ e

−t

= 1 + t +

+ ··· +



1 − t +

−

+ ···



= 2

1 +

+ ···

= 2



n=0

(2n)!

… 2



n=0

/2)

since (2n)! Ú n!2

= 2e

Therefore,

E[e

] Ú e

Since the moment generating function of the sum of independent random variables

is the product of their moment generating functions, we have

E[e

] = (E[e

])

… e

Using the preceding result along with the Chernoff bound gives

P{S

Ú a} … e

−ta

t > 0

The value of t that minimizes the right side of the preceding is the value that min-

imizes nt

/2 − ta, and this value is t = a/n. Supposing that a > 0 (so that the

minimizing t is positive) and letting t = a/n in the preceding inequality yields

P{S

Ú a} … e

−a

/2n

a > 0

This latter inequality yields, for example,

P{S

Ú 6} … e

−36/20

L .1653

Section 8.5 Other Inequalities 409

whereas the exact probability is

P{S

Ú 6}=P{gambler wins at least 8 of the ﬁrst 10 games}













1024

L .0547 .

The next inequality is one having to do with expectations rather than probabilities.

Before stating it, we need the following deﬁnition.

Deﬁnition

A twice-differentiable real-valued function f (x) is said to be convex if f



(x) Ú 0for

all x; similarly, it is said to be concave if f



(x) … 0.

Some examples of convex functions are f(x) = x

, f (x) = e

,andf (x) =−x

1/n

for

x Ú 0. If f (x) is convex, then g(x) =−f (x) is concave, and vice versa.

Proposition 5.3. Jensen’s inequality

If f (x) is a convex function, then

E[ f (X)] Ú f (E[X])

provided that the expectations exist and are ﬁnite.

Proof. Expanding f (x) in a Taylor’s series expansion about μ = E[X] yields

f (x) = f (μ) + f



(μ)(x − μ) +



(ξ)(x − μ)

where ξ is some value between x and μ. Since f



(ξ) Ú 0, we obtain

f (x) Ú f (μ) + f



(μ)(x − μ)

Hence,

f (X) Ú f (μ) + f



(μ)(X − μ)

Taking expectations yields

E[ f (X)] Ú f (μ) + f



(μ)E[X − μ] = f (μ)

and the inequality is established.

EXAMPLE 5f

An investor is faced with the following choices: Either she can invest all of her money

in a risky proposition that would lead to a random return X that has mean m,or

she can put the money into a risk-free venture that will lead to a return of m with

probability 1. Suppose that her decision will be made on the basis of maximizing the

expected value of u(R), where R is her return and u is her utility function. By Jensen’s

inequality, it follows that if u is a concave function, then E[u(X)] … u(m), so the risk-

free alternative is preferable, whereas if u is convex, then E[u(X)] Ú u(m),sothe

risky investment alternative would be preferred. .

410 Chapter 8 Limit Theorems

8.6 BOUNDING THE ERROR PROBABILITY WHEN APPROXIMATING A SUM OF INDEPENDENT

BERNOULLI RANDOM VARIABLES BY A POISSON RANDOM VARIABLE

In this section, we establish bounds on how closely a sum of independent Bernoulli

random variables is approximated by a Poisson random variable with the same mean.

Suppose that we want to approximate the sum of independent Bernoulli random

variables with respective means p

, p

, ..., p

. Starting with a sequence Y

, ..., Y

of independent Poisson random variables, with Y

having mean p

, we will construct

a sequence of independent Bernoulli random variables X

, ..., X

with parameters

, ..., p

such that

P{X

Z Y

} … p

for each i

Letting X =



i=1

and Y =



i=1

, we will use the preceding inequality to con-

clude that

P{X Z Y} …



i=1

Finally, we will show that the preceding inequality implies that, for any set of real

numbers A,

|P{X ∈ A}−P{Y ∈ A}| …



i=1

Since X is the sum of independent Bernoulli random variables and Y is a Poisson

random variable, the latter inequality will yield the desired bound.

To show how the task is accomplished, let Y

, i = 1, ..., n be independent Pois-

son random variables with respective means p

. Now let U

, ..., U

be independent

random variables that are also independent of the Y

’s and which are such that

0 with probability (1 − p

1 with probability 1 − (1 − p

This deﬁnition implicitly makes use of the inequality

−p

Ú 1 − p

in assuming that (1 − p

… 1.

Next, deﬁne the random variables X

, i = 1, ..., n,by

0ifY

= U

= 0

1 otherwise

Note that

P{X

= 0}=P{Y

= 0}P{U

= 0}=e

−p

(1 − p

= 1 − p

P{X

= 1}=1 − P{X

= 0}=p

Section 8.6 Bounding the Error Probability 411

Now, if X

is equal to 0, then so must Y

equal 0 (by the deﬁnition of X

). Therefore,

P{X

Z Y

}=P{X

= 1, Y

Z 1}

= P{Y

= 0, X

= 1}+P{Y

> 1}

= P{Y

= 0, U

= 1}+P{Y

> 1}

= e

−p

[1 − (1 − p

] + 1 − e

−p

− p

−p

= p

− p

−p

… p

(since 1 − e

−p

… p)

Now let X =



i=1

and Y =



i=1

, and note that X is the sum of independent

Bernoulli random variables and Y is Poisson with the expected value E[Y] = E[X] =



i=1

. Note also that the inequality X Z Y implies that X

Z Y

for some i,so

P{X Z Y} … P{X

Z Y

for some i}

…



i=1

P{X

Z Y

} (Boole’s inequality)

…



i=1

For any event B, let I

, the indicator variable for the event B,bedeﬁnedby

1ifB occurs

0 otherwise

Note that, for any set of real numbers A,

{X∈A}

− I

{Y∈A}

… I

{XZY}

The preceding inequality follows from the fact that, since an indicator variable is

either 0 or 1, the left-hand side equals 1 only when I

{X∈A}

= 1andI

{Y∈A}

= 0. But

this would imply that X ∈ A and Y ∈ A, which means that X Z Y, so the right side

would also equal 1. Upon taking expectations of the preceding inequality, we obtain

P{X ∈ A}−P{Y ∈ A} … P{X Z Y}

By reversing X and Y, we obtain, in the same manner,

P{Y ∈ A}−P{X ∈ A} … P{X Z Y}

Thus, we can conclude that

|P{X ∈ A}−P{Y ∈ A}| … P{X Z Y}

Therefore, we have proven that with λ =



i=1

⎧

⎨

⎩



i=1

∈ A

⎫

⎬

⎭

−



i∈A

−λ

…



i=1

412 Chapter 8 Limit Theorems

Remark. When all the p

are equal to p, X is a binomial random variable. Hence,

the preceding inequality shows that, for any set of nonnegative integers A,



i∈A





(1 − p)

n−i

−



i∈A

−np

(np)

… np

SUMMARY

Two useful probability bounds are provided by the Markov and Chebyshev inequali-

ties. The Markov inequality is concerned with nonnegative random variables and says

that, for X of that type,

P{X Ú a} …

E[X]

for every positive value a. The Chebyshev inequality, which is a simple consequence

of the Markov inequality, states that if X has mean μ and variance σ

, then, for every

positive k,

P{|X − μ| Ú kσ } …

The two most important theoretical results in probability are the central limit theorem

and the strong law of large numbers. Both are concerned with a sequence of indepen-

dent and identically distributed random variables. The central limit theorem says that

if the random variables have a ﬁnite mean μ and a ﬁnite variance σ

, then the distri-

bution of the sum of the ﬁrst n of them is, for large n, approximately that of a normal

random variable with mean nμ and variance nσ

. That is, if X

, i Ú 1, is the sequence,

then the central limit theorem states that, for every real number a,

lim

n→q

+ ··· + X

− nμ

√

… a

√

2π

−q

−x

The strong law of large numbers requires only that the random variables in the sequence

have a ﬁnite mean μ. It states that, with probability 1, the average of the ﬁrst n of them

will converge to μ as n goes to inﬁnity. This implies that if A is any speciﬁed event

of an experiment for which independent replications are performed, then the limit-

ing proportion of experiments whose outcomes are in A will, with probability 1, equal

P(A). Therefore, if we accept the interpretation that “with probability 1” means “with

certainty,” we obtain the theoretical justiﬁcation for the long-run relative frequency

interpretation of probabilities.

PROBLEMS

8.1. Suppose that X is a random variable with mean

and variance both equal to 20. What can be said

about P{0 < X < 40}?

8.2. From past experience, a professor knows that the

test score of a student taking her ﬁnal examination

is a random variable with mean 75.

(a) Give an upper bound for the probability that

a student’s test score will exceed 85. Sup-

pose, in addition, that the professor knows

that the variance of a student’s test score is

equal to 25.

(b) What can be said about the probability that a

student will score between 65 and 85?

examination to ensure, with probability at

least .9, that the class average would be

within 5 of 75? Do not use the central limit

theorem.

Problems 413

8.3. Use the central limit theorem to solve part (c) of

Problem 2.

8.4. Let X

, ..., X

be independent Poisson random

variables with mean 1.

(a) Use the Markov inequality to obtain a

bound on

⎧

⎨

⎩



> 15

⎫

⎬

⎭

(b) Use the central limit theorem to approximate

⎧

⎨

⎩



> 15

⎫

⎬

⎭

8.5. Fifty numbers are rounded off to the nearest inte-

ger and then summed. If the individual round-

off errors are uniformly distributed over (−.5, .5),

approximate the probability that the resultant sum

differs from the exact sum by more than 3.

8.6. A die is continually rolled until the total sum of

all rolls exceeds 300. Approximate the probability

that at least 80 rolls are necessary.

8.7. A person has 100 light bulbs whose lifetimes are

independent exponentials with mean 5 hours. If

the bulbs are used one at a time, with a failed bulb

being replaced immediately by a new one, approx-

imate the probability that there is still a working

bulb after 525 hours.

8.8. In Problem 7, suppose that it takes a random time,

uniformly distributed over (0, .5), to replace a

failed bulb. Approximate the probability that all

bulbs have failed by time 550.

8.9. If X is a gamma random variable with parameters

(n, 1), approximately how large need n be so that

− 1

>.01

<.01?

8.10. Civil engineers believe that W, the amount of

weight (in units of 1000 pounds) that a certain span

of a bridge can withstand without structural dam-

age resulting, is normally distributed with mean

400 and standard deviation 40. Suppose that the

weight (again, in units of 1000 pounds) of a car

is a random variable with mean 3 and standard

deviation .3. Approximately how many cars would

have to be on the bridge span for the probability

of structural damage to exceed .1?

8.11. Many people believe that the daily change of price

of a company’s stock on the stock market is a ran-

dom variable with mean 0 and variance σ

.That

is, if Y

represents the price of the stock on the nth

day, then

= Y

n−1

+ X

n Ú 1

where X

, X

, ... are independent and identically

distributed random variables with mean 0 and

variance σ

. Suppose that the stock’s price today

is 100. If σ

= 1, what can you say about the prob-

ability that the stock’s price will exceed 105 after

10 days?

8.12. We have 100 components that we will put in use in

a sequential fashion. That is, component 1 is ini-

tially put in use, and upon failure, it is replaced

by component 2, which is itself replaced upon fail-

ure by component 3, and so on. If the lifetime

of component i is exponentially distributed with

mean 10 + i/10, i = 1, ..., 100, estimate the prob-

ability that the total life of all components will

exceed 1200. Now repeat when the life distribu-

tion of component i is uniformly distributed over

(0, 20 + i/5), i = 1, ..., 100.

8.13. Student scores on exams given by a certain instruc-

tor have mean 74 and standard deviation 14. This

instructor is about to give two exams, one to a class

of size 25 and the other to a class of size 64.

(a) Approximate the probability that the average

test score in the class of size 25 exceeds 80.

(b) Repeat part (a) for the class of size 64.

test score in the larger class exceeds that of

the other class by over 2.2 points.

(d) Approximate the probability that the average

test score in the smaller class exceeds that of

the other class by over 2.2 points.

8.14. A certain component is critical to the operation of

an electrical system and must be replaced immedi-

ately upon failure. If the mean lifetime of this type

of component is 100 hours and its standard devi-

ation is 30 hours, how many of these components

must be in stock so that the probability that the

system is in continual operation for the next 2000

hours is at least .95?

8.15. An insurance company has 10,000 automobile pol-

icyholders. The expected yearly claim per policy-

holder is $240, with a standard deviation of $800.

Approximate the probability that the total yearly

claim exceeds $2.7 million.

8.16. A.J. has 20 jobs that she must do in sequence, with

the times required to do each of these jobs being

independent random variables with mean 50 min-

utes and standard deviation 10 minutes. M.J. has

20 jobs that he must do in sequence, with the times

required to do each of these jobs being indepen-

dent random variables with mean 52 minutes and

standard deviation 15 minutes.

(a) Find the probability that A.J. ﬁnishes in less

than 900 minutes.

(b) Find the probability that M.J. ﬁnishes in less

than 900 minutes.

before M.J.

414 Chapter 8 Limit Theorems

8.17. Redo Example 5b under the assumption that the

number of man–woman pairs is (approximately)

normally distributed. Does this seem like a reason-

able supposition?

8.18. Repeat part (a) of Problem 2 when it is known that

the variance of a student’s test score is equal to 25.

8.19. A lake contains 4 distinct types of ﬁsh. Suppose

that each ﬁsh caught is equally likely to be any

one of these types. Let Y denote the number of

ﬁsh that need be caught to obtain at least one of

each type.

(a) Give an interval (a, b) such that P{a … Y … b}

Ú .90.

(b) Using the one-sided Chebyshev inequality,

how many ﬁsh need we plan on catching so as

to be at least 90 percent certain of obtaining

at least one of each type.

8.20. If X is a nonnegative random variable with mean

25, what can be said about

(a) E[X

(b) E[

√

X]?

(d) E[e

−X

8.21. Let X be a nonnegative random variable.

Prove that

E[X] … (E[X

])

1/2

… (E[X

])

1/3

… ···

8.22. Would the results of Example 5f change if the

investor were allowed to divide her money and

invest the fraction α,0 <α<1, in the risky propo-

sition and invest the remainder in the risk-free

venture? Her return for such a split investment

would be R = αX + (1 − α)m.

8.23. Let X be a Poisson random variable with mean 20.

(a) Use the Markov inequality to obtain an upper

bound on

p = P{X Ú 26}

(b) Use the one-sided Chebyshev inequality to

obtain an upper bound on p.

bound on p.

(d) Approximate p by making use of the central

limit theorem.

(e) Determine p by running an appropriate pro-

gram.

THEORETICAL EXERCISES

8.1. If X has variance σ

,thenσ , the positive square

root of the variance, is called the standard devia-

tion.IfX has mean μ and standard deviation σ ,

show that

P{|X − μ| Ú kσ } …

8.2. If X has mean μ and standard deviation σ ,the

ratio r K |μ|/σ is called the measurement signal-

to-noise ratio of X. The idea is that X can be

expressed as X = μ + (X − μ), with μ represent-

ing the signal and X − μ the noise. If we deﬁne

|(X − μ)/μ| K D as the relative deviation of X

from its signal (or mean) μ, show that, for α>0,

P{D … α} Ú 1 −

8.3. Compute the measurement signal-to-noise ratio—

that is, |μ|/σ,whereμ = E[X]andσ

= Var(X)—

of the following random variables:

(a) Poisson with mean λ;

(b) binomial with parameters n and p;

(d) uniform over (a, b);

(e) exponential with mean 1/λ;

(f) normal with parameters μ, σ

8.4. Let Z

, n Ú 1, be a sequence of random variables

and c a constant such that, for each ε>0, P{|Z

−

c| >ε}

→0asn →q. Show that, for any bounded

continuous function g,

E[g(Z

)] →g(c) as n→q

8.5. Let f(x) be a continuous function deﬁned for 0 …

x … 1. Consider the functions

(x) =



k=0







(1 − x)

n−k

(called Bernstein polynomials) and prove that

lim

n→q

(x) = f (x)

Hint:LetX

, X

, ... be independent Bernoulli

random variables with mean x. Show that

(x) = E





+ ··· + X





and then use Theoretical Exercise 4.

Since it can be shown that the convergence of

(x) to f (x) is uniform in x, the preceding rea-

soning provides a probabilistic proof of the famous

Weierstrass theorem of analysis, which states that

Self-Test Problems and Exercises 415

any continuous function on a closed interval can be

approximated arbitrarily closely by a polynomial.

8.6. (a) Let X be a discrete random variable whose

possible values are 1, 2, ....IfP{X = k} is

nonincreasing in k = 1, 2, ..., prove that

P{X = k} … 2

E[X]

(b) Let X be a nonnegative continuous random

variable having a nonincreasing density func-

tion. Show that

f (x) …

2E[X]

for all x > 0

8.7. Suppose that a fair die is rolled 100 times. Let X

be the value obtained on the ith roll. Compute an

approximation for

⎧

⎨

⎩

100



… a

100

⎫

⎬

⎭

1 < a < 6

8.8. Explain why a gamma random variable with

parameters (t, λ) has an approximately normal dis-

tribution when t is large.

8.9. Suppose a fair coin is tossed 1000 times. If the ﬁrst

100 tosses all result in heads, what proportion of

heads would you expect on the ﬁnal 900 tosses?

Comment on the statement “The strong law of

large numbers swamps, but does not compensate.”

8.10. If X is a Poisson random variable with mean λ,

show that for i <λ,

P{X … i} …

−λ

(eλ)

8.11. Let X be a binomial random variable with param-

eters n and p. Show that, for i > np,

(a) minimum

t>0

−ti

E[e

] occurs when t is such that

(n−i)p

,whereq = 1 − p.

(b) P{X Ú i} …

(n−i)

n−i

(1 − p)

n−i

8.12. The Chernoff bound on a standard normal random

variable Z gives P{Z > a} … e

−a

, a > 0. Show,

by considering the density of Z, that the right side

of the inequality can be reduced by the factor 2.

That is, show that

P{Z > a} …

−a

a > 0

8.13. Show that if E[X] < 0andθ Z 0 is such that

E[e

θX

] = 1, then θ>0.

SELF-TEST PROBLEMS AND EXERCISES

8.1. The number of automobiles sold weekly at a cer-

tain dealership is a random variable with expected

value 16. Give an upper bound to the probabil-

ity that

(a) next week’s sales exceed 18;

(b) next week’s sales exceed 25.

8.2. Suppose in Problem 1 that the variance of the

number of automobiles sold weekly is 9.

(a) Give a lower bound to the probability that

next week’s sales are between 10 and 22,

inclusively.

(b) Give an upper bound to the probability that

next week’s sales exceed 18.

8.3. If

E[X] = 75 E[Y] = 75 Var(X) = 10

Var(Y) = 12 Cov(X, Y) =−3

give an upper bound to

(a) P{|X − Y| > 15};

(b) P{X > Y + 15};

8.4. Suppose that the number of units produced daily

at factory A is a random variable with mean 20 and

standard deviation 3 and the number produced

at factory B is a random variable with mean

18 and standard deviation 6. Assuming indepen-

dence, derive an upper bound for the probability

that more units are produced today at factory B

than at factory A.

8.5. The amount of time that a certain type of compo-

nent functions before failing is a random variable

with probability density function

f (x) = 2x 0 < x < 1

Once the component fails, it is immediately

replaced by another one of the same type. If we

let X

denote the lifetime of the ith component to

be put in use, then S



i=1

represents the time

of the nth failure. The long-term rate at which fail-

ures occur, call it r, is deﬁned by

r = lim

n→q

Assuming that the random variables X

, i Ú 1, are

independent, determine r.

8.6. In Self-Test Problem 5, how many compo-

nents would one need to have on hand to be

416 Chapter 8 Limit Theorems

approximately 90 percent certain that the stock

will last at least 35 days?

8.7. The servicing of a machine requires two separate

steps, with the time needed for the ﬁrst step being

an exponential random variable with mean .2 hour

and the time for the second step being an inde-

pendent exponential random variable with mean

.3 hour. If a repair person has 20 machines to ser-

vice, approximate the probability that all the work

can be completed in 8 hours.

8.8. On each bet, a gambler loses 1 with probability .7,

loses 2 with probability .2, or wins 10 with prob-

ability .1. Approximate the probability that the

gambler will be losing after his ﬁrst 100 bets.

8.9. Determine t so that the probability that the repair

person in Self-Test Problem 7 ﬁnishes the 20 jobs

within time t is approximately equal to .95.

8.10. A tobacco company claims that the amount of

nicotine in one of its cigarettes is a random

variable with mean 2.2 mg and standard devia-

tion .3 mg. However, the average nicotine con-

tent of 100 randomly chosen cigarettes was 3.1

mg. Approximate the probability that the average

would have been as high as or higher than 3.1 if the

company’s claims were true.

8.11. Each of the batteries in a collection of 40 batter-

ies is equally likely to be either a type A or a type

B battery. Type A batteries last for an amount of

time that has mean 50 and standard deviation 15;

type B batteries last for an amount of time that has

mean 30 and standard deviation 6.

(a) Approximate the probability that the total life

of all 40 batteries exceeds 1700.

(b) Suppose it is known that 20 of the batteries

are type A and 20 are type B. Now approx-

imate the probability that the total life of all

40 batteries exceeds 1700.

8.12. A clinic is equally likely to have 2, 3, or 4 doctors

volunteer for service on a given day. No matter

how may volunteer doctors there are on a given

day, the numbers of patients seen by these doctors

are independent Poisson random variables with

mean 30. Let X denote the number of patients

seen in the clinic on a given day.

(a) Find E[X].

(b) Find Var(X).

distribution to approximate P{X > 65}.

8.13. The strong law of large numbers states that, with

probability 1, the successive arithmetic averages

of a sequence of independent and identically dis-

tributed random variables converge to their com-

mon mean μ. What do the successive geometric

averages converge to? That is, what is

lim

n→q

(



i=1

)

1/n

CHAPTER 9

Additional Topics in Probability

9.1 THE POISSON PROCESS

9.2 MARKOV CHAINS

9.3 SURPRISE, UNCERTAINTY, AND ENTROPY

9.4 CODING THEORY AND ENTROPY

9.1 THE POISSON PROCESS

Before we deﬁne a Poisson process, let us recall that a function f is said to be o(h) if

lim

h→0

f (h)

= 0.

That is, f is o(h) if, for small values of h, f (h) is small even in relation to h. Sup-

pose now that “events” are occurring at random points at time, and let N(t) denote

the number of events that occur in the time interval [0, t]. The collection of random

variables {N(t), t Ú 0} is said to be a Poisson process having rate λ, λ>0, if

(i) N(0) = 0.

(ii) The numbers of events that occur in disjoint time intervals are independent.

(iii) The distribution of the number of events that occur in a given interval depends

only on the length of that interval and not on its location.

(iv) P{N(h) = 1}=λh + o(h).

(v) P{N(h) Ú 2}=o(h).

Thus, condition (i) states that the process begins at time 0. Condition (ii), the inde-

pendent increment assumption, states, for instance, that the number of events that

occur by time t [that is, N(t)

] is independent of the number of events that occur

between t and t + s [that is, N(t + s) − N(t)]. Condition (iii), the stationary increment

assumption, states that the probability distribution of N(t + s) − N(t) is the same for

all values of t.

In Chapter 4, we presented an argument, based on the Poisson distribution being

a limiting version of the binomial distribution, that the foregoing conditions imply

that N(t) has a Poisson distribution with mean λt. We will now obtain this result by a

different method.

Lemma 1.1

For a Poisson process with rate λ,

P{N(t) = 0}=e

−λt

417

418 Chapter 9 Additional Topics in Probability

Proof. Let P

(t) = P{N(t) = 0}. We derive a differential equation for P

(t) in the

following manner:

(t + h) = P{N(t + h) = 0}

= P{N(t) = 0, N(t + h) − N(t) = 0}

= P{N(t) = 0}P{N(t + h) − N(t) = 0}

= P

(t)[1 − λh + o(h)]

where the ﬁnal two equations follow from condition (ii) plus the fact that condi-

tions (iv) and (v) imply that P{N(h) = 0}=1 − λh + o(h). Hence,

(t + h) − P

(t)

=−λP

(t) +

o(h)

Now, letting h

→0, we obtain



(t) =−λP

(t)

or, equivalently,



(t)

=−λ

which implies, by integration, that

log P

(t) =−λt + c

(t) = Ke

−λt

Since P

(0) = P{N(0) = 0}=1, we arrive at

(t) = e

−λt

For a Poisson process, let T

denote the time the ﬁrst event occurs. Further, for

n > 1, let T

denote the time elapsed between the (n − 1)st and the nth event. The

sequence {T

, n = 1, 2, ...} is called the sequence of interarrival times. For instance, if

= 5andT

= 10, then the ﬁrst event of the Poisson process would have occurred

at time 5 and the second at time 15.

We shall now determine the distribution of the T

. To do so, we ﬁrst note that the

event {T

> t} takes place if and only if no events of the Poisson process occur in the

interval [0, t]; thus,

P{T

> t}=P{N(t) = 0}=e

−λt

Hence, T

has an exponential distribution with mean 1/λ. Now,

P{T

> t}=E[P{T

> t|T

}]

However,

P{T

> t|T

= s}=P{0 events in (s, s + t]|T

= s}

= P{0 events in (s, s + t]}

= e

−λt

Section 9.2 Markov Chains 419

where the last two equations followed from the assumptions about independent and

stationary increments. From the preceding, we conclude that T

is also an exponen-

tial random variable with mean 1/λ and, furthermore, that T

is independent of T

Repeating the same argument yields Proposition 1.1.

Proposition 1.1. T

, T

, ...are independent exponential random variables, each with

mean 1/λ.

Another quantity of interest is S

, the arrival time of the nth event, also called the

waiting time until the nth event. It is easily seen that



i=1

n Ú 1

hence, from Proposition 1.1 and the results of Section 5.6.1, it follows that S

has a

gamma distribution with parameters n and λ. That is, the probability density of S

given by

(x) = λe

−λx

(λx)

n−1

(n − 1)!

x Ú 0

We are now ready to prove that N(t) is a Poisson random variable with mean λt.

Theorem 1.1. For a Poisson process with rate λ,

P{N(t) = n}=

−λt

(λt)

Proof. Note that the nth event of the Poisson process will occur before or at time

t if and only if the number of events that occur by t is at least n. That is,

N(t) Ú n 3 S

… t

P{N(t) = n}=P{N(t) Ú n}−P{N(t) Ú n + 1}

= P{S

… t}−P{S

n+1

… t}

λe

−λx

(λx)

n−1

(n − 1)!

dx −

λe

−λx

(λx)

But the integration-by-parts formula

udv = uv −

vdu with u = e

−λx

and

dv = λ[(λx)

n−1

/(n − 1)!] dx yields

λe

−λx

(λx)

n−1

(n − 1)!

dx = e

−λt

(λt)

λe

−λx

(λx)

which completes the proof.

9.2 MARKOV CHAINS

Consider a sequence of random variables X

, X

, ..., and suppose that the set of pos-

sible values of these random variables is {0, 1, ..., M}. It will be helpful to interpret

as being the state of some system at time n, and, in accordance with this interpre-

tation, we say that the system is in state i at time n if X

= i. The sequence of random

variables is said to form a Markov chain if, each time the system is in state i, there is

420 Chapter 9 Additional Topics in Probability

some ﬁxed probability—call it P

—that the system will next be in state j. That is, for

all i

, ..., i

n−1

, i, j,

P{X

n+1

= j|X

= i, X

n−1

= i

n−1

, ..., X

= i

, X

= i

}=P

The values P

,0 … i … M,0 … j … N, are called the transition probabilities of the

Markov chain, and they satisfy

Ú 0



j=0

= 1 i = 0, 1, ..., M

(Why?) It is convenient to arrange the transition probabilities P

in a square array as

follows:

··· P

Such an array is called a matrix.

Knowledge of the transition probability matrix and of the distribution of X

enables

us, in theory, to compute all probabilities of interest. For instance, the joint probabil-

ity mass function of X

, ..., X

is given by

P{X

= i

, X

n−1

= i

n−1

, ..., X

= i

, X

= i

}

= P{X

n−1

= i

n−1

, ..., X

}P{X

n−1

= i

n−1

, ..., X

= i

}

= P

n−1

, i

P{X

n−1

= i

n−1

, ..., X

= i

}

and continual repetition of this argument demonstrates that the preceding is equal to

n−1

, i

n−2

, i

n−1

···P

, i

P{X

= i

}

EXAMPLE 2a

Suppose that whether it rains tomorrow depends on previous weather conditions only

through whether it is raining today. Suppose further that if it is raining today, then it

will rain tomorrow with probability α, and if it is not raining today, then it will rain

tomorrow with probability β.

If we say that the system is in state 0 when it rains and state 1 when it does not,

then the preceding system is a two-state Markov chain having transition probability

matrix

α 1 − α

β 1 − β

That is, P

= α = 1 − P

, P

= β = 1 − P

. .

EXAMPLE 2b

Consider a gambler who either wins 1 unit with probability p or loses 1 unit with

probability 1 − p at each play of the game. If we suppose that the gambler will quit

Section 9.2 Markov Chains 421

playing when his fortune hits either 0 or M, then the gambler’s sequence of fortunes

is a Markov chain having transition probabilities

i,i+1

= p = 1 − P

i,i−1

i = 1, ..., M − 1

= P

= 1 .

EXAMPLE 2c

The husband-and-wife physicists Paul and Tatyana Ehrenfest considered a concep-

tual model for the movement of molecules in which M molecules are distributed

among 2 urns. At each time point one of the molecules is chosen at random and

is removed from its urn and placed in the other one. If we let X

denote the number

of molecules in the ﬁrst urn immediately after the nth exchange, then {X

, X

, ...} is

a Markov chain with transition probabilities

i,i+1

M − i

0 … i … M

i,i−1

0 … i … M

= 0if|j − i| > 1 .

Thus, for a Markov chain, P

represents the probability that a system in state i

will enter state j at the next transition. We can also deﬁne the two-stage transition

probability P

(2)

that a system presently in state i will be in state j after two additional

transitions. That is,

(2)

= P{X

m+2

= j|X

= i}

The P

(2)

can be computed from the P

as follows:

(2)

= P{X

= j|X

= i}



k=0

P{X

= j, X

= k|X

= i}



k=0

P{X

= j|X

= k, X

= i}P{X

= k|X

= i}



k=0

In general, we deﬁne the n-stage transition probabilities, denoted as P

(n)

,by

(n)

= P{X

n+m

= j|X

= i}

Proposition 2.1, known as the Chapman–Kolmogorov equations, shows how the P

(n)

can be computed.

Proposition 2.1. The Chapman–Kolmogorov equations

(n)



k=0

(r)

(n−r)

for all 0 < r < n

422 Chapter 9 Additional Topics in Probability

Proof.

(n)

= P{X

= j|X

= i}



P{X

= j, X

= k|X

= i}



P{X

= j|X

= k, X

= i}P{X

= k|X

= i}



(n−r)

(r)

EXAMPLE 2d A random walk

An example of a Markov chain having a countably inﬁnite state space is the random

walk, which tracks a particle as it moves along a one-dimensional axis. Suppose that

at each point in time the particle will move either one step to the right or one step to

the left with respective probabilities p and 1 − p. That is, suppose the particle’s path

follows a Markov chain with transition probabilities

i, i+1

= p = 1 − P

i, i−1

i = 0, ;1, ...

If the particle is at state i, then the probability that it will be at state j after n transitions

is the probability that (n − i + j)/2 of these steps are to the right and n − [(n − i +

j)/2] = (n + i − j)/2 are to the left. Since each step will be to the right, independently

of the other steps, with probability p, it follows that the above is just the binomial

probability



(n − i + j)/2



(n−i+j)/2

(1 − p)

(n+i−j)/2

where





is taken to equal 0 when x is not a nonnegative integer less than or equal

to n. The preceding formula can be rewritten as

i,i+2k



n + k



n+k

(1 − p)

n−k

k = 0, ;1, ..., ;n

2n+1

i,i+2k+1



2n + 1

n + k + 1



n+k+1

(1 − p)

n−k

k = 0, ;1, ..., ;n, −(n + 1) .

Although the P

(n)

denote conditional probabilities, we can use them to derive

expressions for unconditional probabilities by conditioning on the initial state. For

instance,

P{X

= j}=



P{X

= j|X

= i}P{X

= i}



(n)

P{X

= i}

For a large number of Markov chains, it turns out that P

(n)

converges, as n→q,toa

value π

that depends only on j. That is, for large values of n, the probability of being

Section 9.2 Markov Chains 423

in state j after n transitions is approximately equal to π

, no matter what the initial

state was. It can be shown that a sufﬁcient condition for a Markov chain to possess

this property is that, for some n > 0,

(n)

> 0 for all i, j = 0, 1, ..., M (2.1)

Markov chains that satisfy Equation (2.1) are said to be ergodic. Since Proposition 2.1

yields

(n+1)



k=0

(n)

it follows, by letting n→q, that, for ergodic chains,



k=0

(2.2)

Furthermore, since 1 =



j=0

(n)

, we also obtain, by letting n→q,



j=0

= 1 (2.3)

In fact, it can be shown that the π

,0 … j … M, are the unique nonnegative solutions

of Equations (2.2) and (2.3). All this is summed up in Theorem 2.1, which we state

without proof.

Theorem 2.1. For an ergodic Markov chain,

= lim

n→q

(n)

exists, and the π

,0 … j … M, are the unique nonnegative solutions of



k=0



j=0

= 1

EXAMPLE 2e

Consider Example 2a, in which we assume that if it rains today, then it will rain tomor-

row with probability α, and if it does not rain today, then it will rain tomorrow with

probability β. From Theorem 2.1, it follows that the limiting probabilities π

and π

of rain and of no rain, respectively, are given by

= απ

+ βπ

= (1 − α)π

+ (1 − β)π

+ π

= 1

424 Chapter 9 Additional Topics in Probability

which yields

1 + β − α

1 − α

1 + β − α

For instance, if α = .6andβ = .3, then the limiting probability of rain on the nth day

is π

. .

The quantity π

is also equal to the long-run proportion of time that the Markov

chain is in state j, j = 0, ..., M. To see intuitively why this might be so, let P

denote

the long-run proportion of time the chain is in state j. (It can be proven using the

strong law of large numbers that, for an ergodic chain, such long-run proportions

exist and are constants.) Now, since the proportion of time the chain is in state k is

, and since, when in state k, the chain goes to state j with probability P

, it follows

that the proportion of time the Markov chain is entering state j from state k is equal

to P

. Summing over all k shows that P

, the proportion of time the Markov chain

is entering state j, satisﬁes



Since clearly it is also true that



= 1

it thus follows, since by Theorem 2.1 the π

, j = 0, ..., M are the unique solution of

the preceding, that P

= π

, j = 0, ..., M. The long-run proportion interpretation of π

is generally valid even when the chain is not ergodic.

EXAMPLE 2f

Suppose in Example 2c that we are interested in the proportion of time that there are

j molecules in urn 1, j = 0, ..., M. By Theorem 2.1, these quantities will be the unique

solution of

= π

j−1

M − j + 1

+ π

j+1

j + 1

j = 1, ..., M

= π

M−1



j=0

= 1

However, as it is easily checked that







j = 0, ..., M

satisfy the preceding equations, it follows that these are the long-run proportions of

time that the Markov chain is in each of the states. (See Problem 11 for an explanation

of how one might have guessed at the foregoing solution.) .

Section 9.3 Surprise, Uncertainty, and Entropy 425

9.3 SURPRISE, UNCERTAINTY, AND ENTROPY

Consider an event E that can occur when an experiment is performed. How surprised

would we be to hear that E does, in fact, occur? It seems reasonable to suppose that

the amount of surprise engendered by the information that E has occurred should

depend on the probability of E. For instance, if the experiment consists of rolling a

pair of dice, then we would not be too surprised to hear that E has occurred when

E represents the event that the sum of the dice is even (and thus has probability

whereas we would certainly be more surprised to hear that E has occurred when E is

the event that the sum of the dice is 12 (and thus has probability

In this section, we attempt to quantify the concept of surprise. To begin, let us

agree to suppose that the surprise one feels upon learning that an event E has occurred

depends only on the probability of E, and let us denote by S(p) the surprise evoked

by the occurrence of an event having probability p. We determine the functional form

of S(p) by ﬁrst agreeing on a set of reasonable conditions that S(p) should satisfy and

then proving that these axioms require that S(p) have a speciﬁed form. We assume

throughout that S(p) is deﬁned for all 0 < p … 1, but is not deﬁned for events having

p = 0.

Our ﬁrst condition is just a statement of the intuitive fact that there is no surprise

in hearing that an event which is sure to occur has indeed occurred.

Axiom 1

S(1) = 0

Our second condition states that the more unlikely an event is to occur, the greater

is the surprise evoked by its occurrence.

Axiom 2

S(p) is a strictly decreasing function of p; that is, if p < q,thenS(p)>S(q).

The third condition is a mathematical statement of the fact that we would intu-

itively expect a small change in p to correspond to a small change in S(p).

Axiom 3

S(p) is a continuous function of p.

To motivate the ﬁnal condition, consider two independent events E and F having

respective probabilities

P(E) = p and P(F) = q. Since P(EF) = pq, the surprise

evoked by the information that both E and F have occurred is S(pq). Now, suppose

that we are told ﬁrst that E has occurred and then, afterward, that F has also occurred.

Since S(p) is the surprise evoked by the occurrence of E, it follows that S(pq) −

S(p) represents the additional surprise evoked when we are informed that F has also

occurred. However, because F is independent of E, the knowledge that E occurred

does not change the probability of F; hence, the additional surprise should just be

S(q). This reasoning suggests the ﬁnal condition.

Axiom 4

S(pq) = S(p) + S(q) 0 < p … 1, 0 < q … 1

We are now ready for Theorem 3.1, which yields the structure of S(p).

Theorem 3.1.

If S(·) satisﬁes Axioms 1 through 4, then

S(p) =−C log

where C is an arbitrary positive integer.

426 Chapter 9 Additional Topics in Probability

Proof. It follows from Axiom 4 that

S(p

) = S(p) + S(p) = 2S(p)

and by induction that

S(p

) = mS(p) (3.1)

Also, since, for any integral n, S(p) = S(p

1/n

···p

1/n

) = nS(p

1/n

), it follows that

S(p

1/n

) =

S(p) (3.2)

Thus, from Equations (3.1) and (3.2), we obtain

S(p

m/n

) = mS(p

1/n

)

S(p)

which is equivalent to

S(p

) = xS(p) (3.3)

whenever x is a positive rational number. But by the continuity of S (Axiom 3), it

follows that Equation (3.3) is valid for all nonnegative x. (Reason this out.)

Now, for any p,0 < p … 1, let x =−log

p. Then p =





, and from Equa-

tion (3.3),

S(p) = S









= xS





=−C log

where C = S





> S(1) = 0byAxioms2and1.

It is usual to let C equal 1, in which case the surprise is said to be expressed in units

of bits (short for binary digits).

Next, consider a random variable X that must take on one of the values x

, ..., x

with respective probabilities p

, ..., p

. Since −log p

represents the surprise evoked

if X takes on the value x

†

it follows that the expected amount of surprise we shall

receive upon learning the value of X is given by

H(X) =−



i=1

log p

The quantity H(X) is known in information theory as the entropy of the random

variable X. (In case one of the p

= 0, we take 0 log 0 to equal 0.) It can be shown

(and we leave it as an exercise) that H(X) is maximized when all of the p

are equal.

(Is this intuitive?)

Since H(X) represents the average amount of surprise one receives upon learning

the value of X, it can also be interpreted as representing the amount of uncertainty

that exists as to the value of X. In fact, in information theory, H(X) is interpreted as

the average amount of information received when the value of X is observed. Thus,

the average surprise evoked by X, the uncertainty of X, or the average amount of

†

For the remainder of this chapter, we write log x for log

x. Also, we use ln x for log

Section 9.3 Surprise, Uncertainty, and Entropy 427

information yielded by X all represent the same concept viewed from three slightly

different points of view.

Now consider two random variables X and Y that take on the respective values

, ..., x

and y

, ..., y

with joint mass function

p(x

, y

) = P{X = x

, Y = y

}

It follows that the uncertainty as to the value of the random vector (X, Y), denoted

by H(X, Y), is given by

H(X, Y) =−



p(x

, y

) log p(x

, y

)

Suppose now that Y is observed to equal y

. In this situation, the amount of uncer-

tainty remaining in X is given by

Y=y

(X) =−



p(x

) log p(x

)

where

p(x

) = P{X = x

|Y = y

}

Hence, the average amount of uncertainty that will remain in X after Y is observed

is given by

(X) =



Y=y

(X)p

)

where

) = P{Y = y

}

Proposition 3.1 relates H(X, Y) to H(Y) and H

(X). It states that the uncertainty as

to the value of X and Y is equal to the uncertainty of Y plus the average uncertainty

remaining in X when Y is to be observed.

Proposition 3.1.

H(X, Y) = H(Y) + H

(X)

Proof. Using the identity p(x

, y

) = p

)p(x

) yields

H(X, Y) =−



p(x

, y

) log p(x

, y

)

=−



)p(x

)[log p

) + log p(x

)]

=−



) log p

)



p(x

)

−



)



p(x

) log p(x

)

= H(Y) + H

(X)

It is a fundamental result in information theory that the amount of uncertainty in

a random variable X will, on the average, decrease when a second random variable

Y is observed. Before proving this statement, we need the following lemma, whose

proof is left as an exercise.

428 Chapter 9 Additional Topics in Probability

Lemma 3.1

ln x … x − 1 x > 0

with equality only at x = 1.

Theorem 3.2.

(X) … H(X)

with equality if and only if X and Y are independent.

Proof.

(X) − H(X) =−



p(x

) log[p(x

)]p(y

)



p(x

, y

) log p(x

)



p(x

, y

) log



p(x

)

p(x

)



… log e



p(x

, y

)



p(x

)

p(x

)

− 1



by Lemma 3.1

= log e

⎡

⎣



p(x

)p(y

) −



p(x

, y

)

⎤

⎦

= log e[1 − 1]

= 0

9.4 CODING THEORY AND ENTROPY

Suppose that the value of a discrete random vector X is to be observed at location

A and then transmitted to location B via a communication network that consists of

two signals, 0 and 1. In order to do this, it is ﬁrst necessary to encode each possible

value of X in terms of a sequence of 0’s and 1’s. To avoid any ambiguity, it is usually

required that no encoded sequence can be obtained from a shorter encoded sequence

by adding more terms to the shorter.

For instance, if X can take on four possible values x

, x

,andx

, then one pos-

sible coding would be

% 00

% 01

% 10

% 11

(4.1)

That is, if X = x

, then the message 00 is sent to location B, whereas if X = x

,then

01 is sent to B, and so on. A second possible coding is

% 0

% 10

% 110

% 111

(4.2)

Section 9.4 Coding Theory and Entropy 429

However, a coding such as

% 0

% 1

% 00

% 01

is not allowed because the coded sequences for x

and x

are both extensions of the

one for x

One of the objectives in devising a code is to minimize the expected number of

bits (that is, binary digits) that need to be sent from location A to location B.For

example, if

P{X = x

then the code given by Equation (4.2) would expect to send

(1) +

(2) +

(3) +

(3) = 1.75 bits, whereas the code given by Equation (4.1) would expect to send 2

bits. Hence, for the preceding set of probabilities, the encoding in Equation (4.2) is

more efﬁcient than that in Equation (4.1).

The preceding discussion raises the following question: For a given random

vector X, what is the maximum efﬁciency achievable by an encoding scheme? The

answer is that, for any coding, the average number of bits that will be sent is at least

as large as the entropy of X. To prove this result, known in information theory as the

noiseless coding theorem, we shall need Lemma 4.1.

Lemma 4.1

Let X take on the possible values x

, ..., x

. Then, in order to be able to encode

the values of X in binary sequences (none of which is an extension of another) of

respective lengths n

, ..., n

, it is necessary and sufﬁcient that



i=1





… 1

Proof. For a ﬁxed set of N positive integers n

, ..., n

, let w

denote the number

of the n

that are equal to j , j = 1, .... For there to be a coding that assigns n

bits to the value x

, i = 1, ..., N, it is clearly necessary that w

… 2. Furthermore,

because no binary sequence is allowed to be an extension of any other, we must

have w

… 2

− 2w

. (This follows because 2

is the number of binary sequences

of length 2, whereas 2w

is the number of sequences that are extensions of the

binary sequence of length 1.) In general, the same reasoning shows that we

must have

… 2

− w

n−1

− w

n−2

− ··· − w

n−1

2 (4.3)

430 Chapter 9 Additional Topics in Probability

for n = 1, .... In fact, a little thought should convince the reader that these con-

ditions are not only necessary, but also sufﬁcient for a code to exist that assigns n

bits to x

, i = 1, ..., N.

Rewriting inequality (4.3) as

+ w

n−1

2 + w

n−2

+ ··· + w

n−1

… 2

n = 1, ...

and dividing by 2

yields the necessary and sufﬁcient conditions, namely,



j=1





… 1 for all n (4.4)

However, because



j=1





is increasing in n, it follows that Equation (4.4) will

be true if and only if



j=1





… 1

The result is now established, since, by the deﬁnition of w

as the number of n

that

equal j, it follows that



j=1







i=1





We are now ready to prove Theorem 4.1.

Theorem 4.1 The noiseless coding theorem

Let X take on the values x

, ..., x

with respective probabilities p(x

), ..., p(x

Then, for any coding of X that assigns n

bits to x



i=1

p(x

) Ú H(X) =−



i=1

p(x

) log p(x

)

Proof. Let P

= p(x

), q

= 2

−n





j=1

−n

, i = 1, ..., N. Then

−



i=1

log





=−log e



i=1





= log e



i=1





… log e



i=1



− 1



by Lemma 3.1

= 0 since



i=1



i=1

= 1

Section 9.4 Coding Theory and Entropy 431

Hence,

−



i=1

log P

… −



i=1

log q



i=1

+ log

⎛

⎜

⎝



j=1

−n

⎞

⎟

⎠

…



i=1

by Lemma 4.1

EXAMPLE 4a

Consider a random variable X with probability mass function

p(x

) =

p(x

) =

p(x

) = p(x

) =

Since

H(X) =−



log



= 1.75

it follows from Theorem 4.1 that there is no more efﬁcient coding scheme than

% 0

% 10

% 110

% 111 .

For most random vectors, there does not exist a coding for which the average

number of bits sent attains the lower bound H(X). However, it is always possible

to devise a code such that the average number of bits is within 1 of H (X). To prove

this, deﬁne n

to be the integer satisfying

−log p(x

) … n

< −log p(x

) + 1

Now,



i=1

−n

…



i=1

log p(x

)



i=1

p(x

) = 1

so, by Lemma 4.1, we can associate sequences of bits having lengths n

with the x

, i =

1, ..., N. The average length of such a sequence,

L =



i=1

p(x

)

432 Chapter 9 Additional Topics in Probability

satisﬁes

−



i=1

p(x

) log p(x

) … L < −



i=1

p(x

) log p(x

) + 1

H(X) … L < H(X) + 1

EXAMPLE 4b

Suppose that 10 independent tosses of a coin having probability p of coming up heads

are made at location A and the result is to be transmitted to location B. The outcome

of this experiment is a random vector X = (X

, ..., X

), where X

is 1 or 0 according

to whether or not the outcome of the ith toss is heads. By the results of this section, it

follows that L, the average number of bits transmitted by any code, satisﬁes

H(X) … L

with

L … H(X) + 1

for at least one code. Now, since the X

are independent, it follows from Proposi-

tion 3.1 and Theorem 3.2 that

H(X) = H(X

, ..., X

) =



i=1

H(X

)

=−10[p log p + (1 − p) log(1 − p)]

If p =

,thenH(X) = 10, and it follows that we can do no better than just encoding

X by its actual value. For example, if the ﬁrst 5 tosses come up heads and the last 5

tails, then the message 1111100000 is transmitted to location B.

However, if p Z

, we can often do better by using a different coding scheme. For

instance, if p =

,then

H(X) =−10



log



= 8.11

Thus, there is an encoding for which the average length of the encoded message is no

greater than 9.11.

One simple coding that is more efﬁcient in this case than the identity code is to

break up (X

, ..., X

) into 5 pairs of 2 random variables each and then, for i =

1, 3, 5, 7, 9, code each of the pairs as follows:

= 0, X

i+1

= 0 % 0

= 0, X

i+1

= 1 % 10

= 1, X

i+1

= 0 % 110

= 1, X

i+1

= 1 % 111

The total message transmitted is the successive encodings of the preceding pairs.

For instance, if the outcome TTTHHTTTTH is observed, then the message 01011

0010 is sent. The average number of bits needed to transmit the message with this

code is

Section 9.4 Coding Theory and Entropy 433







+ 2







+ 3







+ 3







135

L 8.44 .

Up to this point, we have assumed that the message sent at location A is received

without error at location B. However, there are always certain errors that can occur

because of random disturbances along the communications channel. Such random

disturbances might lead, for example, to the message 00101101, sent at A, being

received at B in the form 01101101.

Let us suppose that a bit transmitted at location A will be correctly received at

location B with probability p, independently from bit to bit. Such a communications

system is called a binary symmetric channel. Suppose further that p = .8 and we

want to transmit a message consisting of a large number of bits from A to B. Thus,

direct transmission of the message will result in an error probability of .20 for each

bit, which is quite high. One way to reduce this probability of bit error would be to

transmit each bit 3 times and then decode by majority rule. That is, we could use the

following scheme:

Encode Decode Encode Decode

0→000

000

001

010

100

⎫

⎪

⎬

⎪

⎭

→01→111

111

110

101

011

⎫

⎪

⎬

⎪

⎭

→1

Note that if no more than one error occurs in transmission, then the bit will be

correctly decoded. Hence, the probability of bit error is reduced to

(.2)

+ 3(.2)

(.8) = .104

a considerable improvement. In fact, it is clear that we can make the probability of

bit error as small as we want by repeating the bit many times and then decoding by

majority rule. For instance, the scheme

Encode Decode

0→string of 17 0’s By majority rule

→string of 17 1’s

will reduce the probability of bit error to below .01.

The problem with this type of encoding scheme is that, although it decreases the

probability of bit error, it does so at the cost of also decreasing the effective rate of

bits sent per signal. (See Table 9.1.)

In fact, at this point it may appear inevitable to the reader that decreasing the

probability of bit error to 0 always results in also decreasing the effective rate at which

bits are transmitted per signal to 0. However, a remarkable result of information

theory known as the noisy coding theorem and due to Claude Shannon demonstrates

that this is not the case. We now state this result as Theorem 4.2.

434 Chapter 9 Additional Topics in Probability

TABLE 9.1: Repetition of Bits Encoding Scheme

Probability of error Rate

(per bit) (bits transmitted per signal)

.20 1

.10 .33





.01 .06





Theorem 4.2 The noisy coding theorem

There is a number C such that, for any value R which is less than C, and for any ε>0,

there exists a coding–decoding scheme that transmits at an average rate of R bits sent

per signal and with an error (per bit) probability of less than ε. The largest such

value of C—call it C

∗

†

—is called the channel capacity, and for the binary symmetric

channel,

∗

= 1 + p log p + (1 − p) log(1 − p)

SUMMARY

The Poisson process having rate λ is a collection of random variables {N(t), t Ú 0} that

relate to an underlying process of randomly occurring events. For instance, N(t) rep-

resents the number of events that occur between times 0 and t. The deﬁning features

of the Poisson process are as follows:

(i) The number of events that occur in disjoint time intervals are independent.

(ii) The distribution of the number of events that occur in an interval depends only

on the length of the interval.

(iii) Events occur one at a time.

(iv) Events occur at rate λ.

It can be shown that N(t) is a Poisson random variable with mean λt. In addition,

if T

, i Ú 1, are the times between the successive events, then they are independent

exponential random variables with rate λ.

A sequence of random variables X

, n Ú 0, each of which takes on one of the

values 0, ..., M, is said to be a Markov chain with transition probabilities P

i, j

if, for

all n, i

, ..., i

, i, j,

P{X

n+1

= j|X

= i, X

n−1

= i

n−1

, ..., X

= i

}=P

i, j

If we interpret X

as the state of some process at time n, then a Markov chain is a

sequence of successive states of a process which has the property that whenever it

enters state i, then, independently of all past states, the next state is j with probability

i,j

, for all states i and j. For many Markov chains, the probability of being in state j at

time n converges to a limiting value that does not depend on the initial state. If we let

, j = 0, ..., M, denote these limiting probabilities, then they are the unique solution

of the equations



i=0

i, j

j = 0, ..., M



j=1

= 1

†

For an entropy interpretation of C

∗

, see Theoretical Exercise 9.18.

Problems and Theoretical Exercises 435

Moreover, π

is equal to the long-run proportion of time that the chain is in state j.

Let X be a random variable that takes on one of n possible values according to the

set of probabilities {p

, ..., p

}. The quantity

H(X) =−



i=1

log

)

is called the entropy of X. It can be interpreted as representing either the average

amount of uncertainty that exists regarding the value of X or the average information

received when X is observed. Entropy has important implications for binary codings

of X.

PROBLEMS AND THEORETICAL EXERCISES

9.1. Customers arrive at a bank at a Poisson rate λ.

Suppose that two customers arrived during the ﬁrst

hour. What is the probability that

(a) both arrived during the ﬁrst 20 minutes?

(b) at least one arrived during the ﬁrst 20 min-

utes?

9.2. Cars cross a certain point in the highway in accor-

dance with a Poisson process with rate λ = 3per

minute. If Al runs blindly across the highway, what

is the probability that he will be uninjured if the

amount of time that it takes him to cross the road

is s seconds? (Assume that if he is on the highway

when a car passes by, then he will be injured.) Do

this exercise for s = 2, 5, 10, 20.

9.3. Suppose that in Problem 9.2 Al is agile enough to

escape from a single car, but if he encounters two

or more cars while attempting to cross the road,

then he is injured. What is the probability that he

will be unhurt if it takes him s seconds to cross? Do

this exercise for s = 5, 10, 20, 30.

9.4. Suppose that 3 white and 3 black balls are dis-

tributed in two urns in such a way that each urn

contains 3 balls. We say that the system is in state

i if the ﬁrst urn contains i white balls, i = 0, 1, 2, 3.

At each stage, 1 ball is drawn from each urn and

the ball drawn from the ﬁrst urn is placed in the

second, and conversely with the ball from the

second urn. Let X

denote the state of the sys-

tem after the nth stage, and compute the transition

probabilities of the Markov chain {X

, n Ú 0}.

9.5. Consider Example 2a. If there is a 50–50 chance of

rain today, compute the probability that it will rain

3 days from now if α = .7andβ = .3.

9.6. Compute the limiting probabilities for the model

of Problem 9.4.

9.7. A transition probability matrix is said to be doubly

stochastic if



i=0

= 1

for all states j = 0, 1, ..., M. Show that such a

Markov chain is ergodic, then

= 1/(M + 1), j =

0, 1, ..., M.

9.8. On any given day, Buffy is either cheerful (c), so-so

(s), or gloomy (g). If she is cheerful today, then she

will be c, s, or g tomorrow with respective probabil-

ities .7, .2, and .1. If she is so-so today, then she will

be c, s, or g tomorrow with respective probabilities

.4, .3, and .3. If she is gloomy today, then Buffy will

be c, s, or g tomorrow with probabilities .2, .4, and

.4. What proportion of time is Buffy cheerful?

9.9. Suppose that whether it rains tomorrow depends

on past weather conditions only through the last

2 days. Speciﬁcally, suppose that if it has rained

yesterday and today, then it will rain tomorrow

with probability .8; if it rained yesterday but not

today, then it will rain tomorrow with probability

.3; if it rained today but not yesterday, then it will

rain tomorrow with probability .4; and if it has not

rained either yesterday or today, then it will rain

tomorrow with probability .2. What proportion of

days does it rain?

9.10. A certain person goes for a run each morning.

When he leaves his house for his run, he is equally

likely to go out either the front or the back door,

and similarly, when he returns, he is equally likely

to go to either the front or the back door. The run-

ner owns 5 pairs of running shoes, which he takes

off after the run at whichever door he happens to

be. If there are no shoes at the door from which he

leaves to go running, he runs barefooted. We are

interested in determining the proportion of time

that he runs barefooted.

(a) Set this problem up as a Markov chain. Give

the states and the transition probabilities.

(b) Determine the proportion of days that he runs

barefooted.

9.11. This problem refers to Example 2f.

(a) Verify that the proposed value of

satisﬁes

the necessary equations.

436 Chapter 9 Additional Topics in Probability

(b) For any given molecule, what do you think is

the (limiting) probability that it is in urn 1?

j Ú 1, is in urn 1 at a very large time would be

(in the limit) independent?

(d) Explain why the limiting probabilities are as

given.

9.12. Determine the entropy of the sum that is obtained

when a pair of fair dice is rolled.

9.13. Prove that if X cantakeonanyofn possible values

with respective probabilities P

, ..., P

,thenH(X)

is maximized when P

= 1/n, i = 1, ..., n. What is

H(X) equal to in this case?

9.14. Apairoffairdiceisrolled.Let

X =

1 if the sum of the dice is 6

0 otherwise

and let Y equal the value of the ﬁrst die. Compute

(a) H(Y),(b)H

(X),and(c)H(X, Y).

9.15. A coin having probability p =

of coming up

heads is ﬂipped 6 times. Compute the entropy of

the outcome of this experiment.

9.16. A random variable can take on any of n possi-

ble values x

, ..., x

with respective probabilities

p(x

), i = 1, ..., n. We shall attempt to determine

the value of X by asking a series of questions,

each of which can be answered “yes” or “no.” For

instance, we may ask “Is X = x

?” or “Is X equal

to either x

or x

?” and so on. What can you

say about the average number of such questions

that you will need to ask to determine the value

of X?

9.17. Show that, for any discrete random variable X and

function f ,

H(f (X)) … H(X)

9.18. In transmitting a bit from location A to location

B,ifweletX denote the value of the bit sent at

location A and Y denote the value received at loca-

tion B,thenH(X) − H

(X) is called the rate of

transmission of information from A to B. The max-

imal rate of transmission, as a function of P{X =

1}=1 − P{X = 0}, is called the channel capac-

ity. Show that, for a binary symmetric channel with

P{Y = 1|X = 1}=P{Y = 0|X = 0}=p,thechan-

nel capacity is attained by the rate of transmission

of information when P{X = 1}=

and its value is

1 + p log p + (1 − p) log(1 − p).

SELF-TEST PROBLEMS AND EXERCISES

9.1. Events occur according to a Poisson process with

rate λ = 3 per hour.

(a) What is the probability that no events occur

between times 8 and 10 in the morning?

(b) What is the expected value of the number of

events that occur between times 8 and 10 in

the morning?

the ﬁfth event after 2

P.M.?

9.2. Customers arrive at a certain retail establishment

according to a Poisson process with rate λ per

hour. Suppose that two customers arrive during

the ﬁrst hour. Find the probability that

(a) both arrived in the ﬁrst 20 minutes;

(b) at least one arrived in the ﬁrst 30 minutes.

9.3. Four out of every ﬁve trucks on the road are fol-

lowed by a car, while one out of every six cars is

followed by a truck. What proportion of vehicles

on the road are trucks?

9.4. A certain town’s weather is classiﬁed each day as

being rainy, sunny, or overcast, but dry. If it is

rainy one day, then it is equally likely to be either

sunny or overcast the following day. If it is not

rainy, then there is one chance in three that the

weather will persist in whatever state it is in for

another day, and if it does change, then it is equally

likely to become either of the other two states. In

the long run, what proportion of days are sunny?

What proportion are rainy?

9.5. Let X be a random variable that takes on 5 pos-

sible values with respective probabilities .35, .2, .2,

.2, and .05. Also, let Y be a random variable that

takes on 5 possible values with respective proba-

bilities .05, .35, .1, .15, and .35.

(a) Show that H(X)>H(Y).

(b) Using the result of Problem 9.13, give an intu-

itive explanation for the preceding inequality.

REFERENCES

Sections 9.1 and 9.2

[1] KEMENY, J., L. SNELL,andA.KNAPP. Denumerable Markov Chains.NewYork:

D. Van Nostrand Company, 1966.

[2] P

ARZEN,E.Stochastic Processes. San Francisco: Holden-Day, Inc., 1962.

References 437

[3] R

OSS,S.M. Introduction to Probability Models, 9th ed. San Diego: Academic Press,

Inc., 2007.

[4] R

OSS,S.M.Stochastic Processes, 2d ed. New York: John Wiley & Sons, Inc., 1996.

Sections 9.3 and 9.4

[5] ABRAMSON,N. Information Theory and Coding. New York: McGraw-Hill Book

Company, 1963.

[6] M

CELIECE,R. Theory of Information and Coding. Reading, MA: Addison-Wesley

Publishing Co., Inc., 1977.

[7] P

ETERSON,W.,andE.WELDON. Error Correcting Codes, 2d ed. Cambridge, MA: The

MIT Press, 1972.

CHAPTER 10

Simulation

10.1 INTRODUCTION

10.2 GENERAL TECHNIQUES FOR SIMULATING CONTINUOUS RANDOM VARIABLES

10.3 SIMULATING FROM DISCRETE DISTRIBUTIONS

10.4 VARIANCE REDUCTION TECHNIQUES

10.1 INTRODUCTION

How can we determine the probability of our winning a game of solitaire?

(By solitaire, we mean any one of the standard solitaire games played with an ordi-

nary deck of 52 playing cards and with some ﬁxed playing strategy.) One possible

approach is to start with the reasonable hypothesis that all (52)! possible arrange-

ments of the deck of cards are equally likely to occur and then attempt to determine

how many of these lead to a win. Unfortunately, there does not appear to be any sys-

tematic method for determining the number of arrangements that lead to a win, and

as (52)! is a rather large number and the only way to determine whether a particular

arrangement leads to a win seems to be by playing the game out, it can be seen that

this approach will not work.

In fact, it might appear that the determination of the probability of winning at

solitaire is mathematically intractable. However, all is not lost, for probability falls

not only within the realm of mathematics, but also within the realm of applied sci-

ence; and, as in all applied sciences, experimentation is a valuable technique. For our

solitaire example, experimentation takes the form of playing a large number of such

games or, better yet, programming a computer to do so. After playing, say, n games,

if we let

1iftheith game results in a win

0 otherwise

then X

, i = 1, ..., n will be independent Bernoulli random variables for which

E[X

] = P{win at solitaire}

Hence, by the strong law of large numbers, we know that



i=1

number of games won

number of games played

will, with probability 1, converge to P{win at solitaire}. That is, by playing a large

number of games, we can use the proportion of games won as an estimate of the prob-

ability of winning. This method of empirically determining probabilities by means of

experimentation is known as simulation.

438

Section 10.1 Introduction 439

In order to use a computer to initiate a simulation study, we must be able to gener-

ate the value of a uniform (0, 1) random variable; such variates are called random

numbers. To generate them, most computers have a built-in subroutine, called a

random-number generator, whose output is a sequence of pseudorandom numbers—

a sequence of numbers that is, for all practical purposes, indistinguishable from a sam-

ple from the uniform (0, 1) distribution. Most random-number generators start with

an initial value X

, called the seed, and then recursively compute values by specifying

positive integers a, c,andm, and then letting

n+1

= (aX

+ c) modulo mnÚ 0 (1.1)

where the foregoing means that aX

+ c is divided by m and the remainder is taken

as the value of X

n+1

. Thus, each X

is either 0, 1, ..., m − 1, and the quantity X

/m is

taken as an approximation to a uniform (0, 1) random variable. It can be shown that,

subject to suitable choices for a, c,andm, Equation (1.1) gives rise to a sequence of

numbers that look as if they were generated from independent uniform (0, 1) random

variables.

As our starting point in simulation, we shall suppose that we can simulate from

the uniform (0, 1) distribution, and we shall use the term random numbers to mean

independent random variables from this distribution.

In the solitaire example, we would need to program a computer to play out the

game starting with a given ordering of the cards. However, since the initial order-

ing is supposed to be equally likely to be any of the (52)! possible permutations, it

is also necessary to be able to generate a random permutation. Using only random

numbers, the following algorithm shows how this can be accomplished. The algorithm

begins by randomly choosing one of the elements and then putting it in position n;it

then randomly chooses among the remaining elements and puts the choice in position

n − 1, and so on. The algorithm efﬁciently makes a random choice among the remain-

ing elements by keeping these elements in an ordered list and then randomly choosing

a position on that list.

EXAMPLE 1a Generating a random permutation

Suppose we are interested in generating a permutation of the integers 1, 2, ..., n such

that all n! possible orderings are equally likely. Then, starting with any initial permu-

tation, we will accomplish this after n − 1 steps, where we interchange the positions

of two of the numbers of the permutation at each step. Throughout, we will keep

track of the permutation by letting X(i), i = 1, ..., n denote the number currently in

position i. The algorithm operates as follows:

1. Consider any arbitrary permutation, and let X(i) denote the element in posi-

tion i, i = 1 ..., n. [For instance, we could take X(i) = i, i = 1, ..., n.]

2. Generate a random variable N

that is equally likely to equal any of the values

1, 2, ..., n.

3. Interchange the values of X(N

) and X(n). The value of X(n) will now remain

ﬁxed. [For instance, suppose that n = 4 and initially X(i) = i, i = 1, 2, 3, 4. If

= 3, then the new permutation is X(1) = 1, X(2) = 2, X(3) = 4, X(4) = 3,

and element 3 will remain in position 4 throughout.]

4. Generate a random variable N

n−1

that is equally likely to be either 1, 2, ...,

n − 1.

5. Interchange the values of X(N

n−1

) and X(n − 1).[IfN

= 1, then the new

permutation is X(1) = 4, X(2) = 2, X(3) = 1, X(4) = 3.]

440 Chapter 10 Simulation

6. Generate N

n−2

, which is equally likely to be either 1, 2, ..., n − 2.

7. Interchange the values of X(N

n−2

) and X(n − 2).[IfN

= 1, then the new

permutation is X(1) = 2, X(2) = 4, X(3) = 1, X(4) = 3, and this is the ﬁnal

permutation.]

8. Generate N

n−3

, and so on. The algorithm continues until N

is generated, and

after the next interchange the resulting permutation is the ﬁnal one.

To implement this algorithm, it is necessary to be able to generate a random vari-

able that is equally likely to be any of the values 1, 2, ..., k. To accomplish this, let

U denote a random number—that is, U is uniformly distributed on (0, 1)—and note

that kU is uniform on (0, k). Hence,

P{i − 1 < kU < i}=

i = 1, ..., k

so if we take N

= [kU] + 1, where [x] is the integer part of x (that is, the largest

integer less than or equal to x), then N

will have the desired distribution.

The algorithm can now be succinctly written as follows:

Step 1. Let X(1), ..., X(n) be any permutation of 1, 2, ..., n. [For instance, we can

set X(i) = i, i = 1, ..., n.]

Step 2. Let I = n.

Step 3. Generate a random number U and set N = [IU] + 1.

Step 4. Interchange the values of X(N) and X(I).

Step 5. Reduce the value of I by 1, and if I > 1, go to step 3.

Step 6. X(1), ..., X(n) is the desired random generated permutation.

The foregoing algorithm for generating a random permutation is extremely useful.

For instance, suppose that a statistician is developing an experiment to compare the

effects of m different treatments on a set of n subjects. He decides to split the subjects

into m different groups of respective sizes n

, n

, ..., n

, where



i=1

= n,with

the members of the ith group to receive treatment i. To eliminate any bias in the

assignment of subjects to treatments (for instance, it would cloud the meaning of the

experimental results if it turned out that all the “best” subjects had been put in the

same group), it is imperative that the assignment of a subject to a given group be done

“at random.” How is this to be accomplished?

†

A simple and efﬁcient procedure is to arbitrarily number the subjects 1 through

n and then generate a random permutation X(1), ..., X(n) of 1, 2, ..., n. Now assign

subjects X(1), X(2), ..., X(n

) to be in group 1, X(n

+ 1), ..., X(n

+ n

) to be in

group 2, and, in general, group j is to consist of subjects numbered X(n

+ n

+···+

j−1

+ k), k = 1, ..., n

. .

10.2 GENERAL TECHNIQUES FOR SIMULATING CONTINUOUS

RANDOM VARIABLES

In this section, we present two general methods for using random numbers to simulate

continuous random variables.

†

Another technique for randomly dividing the subjects when m = 2 was presented in Exam-

ple 2g of Chapter 6. The preceding procedure is faster, but requires more space than the one of

Example 2g.

Section 10.2 Simulating Continuous Random Variables 441

10.2.1 The Inverse Transformation Method

A general method for simulating a random variable having a continuous distribution—

called the inverse transformation method—is based on the following proposition.

Proposition 2.1. Let U be a uniform (0, 1) random variable. For any continuous

distribution function F, if we deﬁne the random variable Y by

Y = F

−1

(U)

then the random variable Y has distribution function F.[F

−1

(x) is deﬁned to equal

that value y for which F(y) = x.]

Proof.

(a) = P{Y … a}

= P{F

−1

(U) … a} (2.1)

Now, since F(x) is a monotone function, it follows that F

−1

(U) … a if and only if

U … F(a). Hence, from Equation (2.1), we have

(a) = P{U … F(a)}

= F(a)

It follows from Proposition 2.1 that we can simulate a random variable X having

a continuous distribution function F by generating a random number U and then

setting X = F

−1

(U).

EXAMPLE 2a Simulating an exponential random variable

If F(x) = 1 − e

−x

,thenF

−1

(u) is that value of x such that

1 − e

−x

= u

x =−log(1 − u)

Hence, if U is a uniform (0, 1) variable, then

−1

(U) =−log(1 − U)

is exponentially distributed with mean 1. Since 1 − U is also uniformly distributed on

(0, 1), it follows that −log U is exponential with mean 1. Since cX is exponential with

mean c when X is exponential with mean 1, it follows that −c log U is exponential

with mean c. .

The results of Example 2a can also be utilized to stimulate a gamma random

variable.

EXAMPLE 2b Simulating a gamma (n, λ) random variable

To simulate from a gamma distribution with parameters (n, λ) when n is an integer,

we use the fact that the sum of n independent exponential random variables, each

having rate λ, has this distribution. Hence, if U

, ..., U

are independent uniform

(0, 1) random variables, then

442 Chapter 10 Simulation

X =−



i=1

log U

=−

log

⎛

⎝



i=1

⎞

⎠

has the desired distribution. .

10.2.2 The Rejection Method

Suppose that we have a method for simulating a random variable having density

function g(x). We can use this method as the basis for simulating from the contin-

uous distribution having density f (x) by simulating Y from g and then accepting the

simulated value with a probability proportional to f (Y)/g(Y).

Speciﬁcally, let c be a constant such that

f (y)

g(y)

… c for all y

We then have the following technique for simulating a random variable having

density f .

Rejection Method

Step 1. Simulate Y having density g and simulate a random number U.

Step 2. If U … f (Y)/cg(Y), set X = Y. Otherwise return to step 1.

The rejection method is expressed pictorially in Figure 10.1. We now prove that it

works.

Generate

Y ⬃ g

Start

Generate a

random number

f(Y)

——––

cg(Y)

⭐

Yes

Set X =

FIGURE 10.1: Rejection method for simulating a random variable X having density function f.

Proposition 2.2. The random variable X generated by the rejection method has den-

sity function f .

Proof. Let X be the value obtained and let N denote the number of necessary

iterations. Then

P{X … x}=P{Y

… x}

= P

Y … x|U …

f (Y)

cg(Y)

Y … x, U …

f (Y)

cg(Y)

where K = P{U … f (Y)/cg(Y)}. Now, by independence, the joint density function

of Y and U is

f (y, u) = g(y) 0 < u < 1

Section 10.2 Simulating Continuous Random Variables 443

so, using the foregoing, we have

P{X … x}=

y … x

0 … u … f (y)/cg(y)

g(y) du dy

−q

f (y)/cg(y)

du g(y) dy

−q

f (y) dy

(2.2)

Letting X approach q and using the fact that f is a density gives

1 =

−q

f (y) dy =

Hence, from Equation (2.2), we obtain

P{X … x}=

−q

f (y) dy

which completes the proof.

Remarks. (a) Note that the way in which we “accept the value Y with prob-

ability f (Y)/cg(Y)” is by generating a random number U and then accepting Y if

U … f (Y)/cg(Y).

(b) Since each iteration will independently result in an accepted value with proba-

bility P{U … f (Y)/cg(Y)}=K = 1/c, it follows that the number of iterations has a

geometric distribution with mean c. .

EXAMPLE 2c Simulating a normal random variable

To simulate a unit normal random variable Z (that is, one with mean 0 and variance

1), note ﬁrst that the absolute value of Z has probability density function

f (x) =

√

2π

−x

0 < x < q (2.3)

We will start by simulating from the preceding density function by using the rejection

method, with g being the exponential density function with mean 1—that is,

g(x) = e

−x

0 < x < q

Now, note that

f (x)

g(x)

exp

−(x

− 2x)

exp

−(x

− 2x + 1)

exp

−(x − 1)

(2.4)

…

444 Chapter 10 Simulation

Hence, we can take c =

2e/π; so, from Equation (2.4),

f (x)

cg(x)

= exp

−(x − 1)

Therefore, using the rejection method, we can simulate the absolute value of a unit

normal random variable as follows:

(a) Generate independent random variables Y and U, Y being exponential with

rate 1 and U being uniform on (0, 1).

(b) If U … exp{−(Y − 1)

/2}, set X = Y. Otherwise, return to (a).

Once we have simulated a random variable X having Equation (2.3) as its density

function, we can then generate a unit normal random variable Z by letting Z be

equally likely to be either X or −X.

In step (b), the value Y is accepted if U … exp{−(Y − 1)

/2}, which is equiva-

lent to −log U Ú (Y − 1)

/2. However, in Example 2a it was shown that −log U is

exponential with rate 1, so steps (a) and (b) are equivalent to



) Generate independent exponentials Y

and Y

, each with rate 1.



)IfY

Ú (Y

− 1)

/2, set X = Y

. Otherwise, return to (a’).

Suppose now that the foregoing results in Y

’s being accepted—so we know that Y

is larger than (Y

− 1)

/2. By how much does the one exceed the other? To answer

this question, recall that Y

is exponential with rate 1; hence, given that it exceeds

some value, the amount by which Y

exceeds (Y

− 1)

/2 [that is, its “additional

life” beyond the time (Y

− 1)

/2] is (by the memoryless property) also exponentially

distributed with rate 1. That is, when we accept step (b



), not only do we obtain X (the

absolute value of a unit normal), but, by computing Y

− (Y

− 1)

/2, we also can

generate an exponential random variable (that is independent of X) having rate 1.

Summing up, then, we have the following algorithm that generates an exponential

with rate 1 and an independent unit normal random variable:

Step 1. Generate Y

, an exponential random variable with rate 1.

Step 2. Generate Y

, an exponential random variable with rate 1.

Step 3. If Y

− (Y

− 1)

/2 > 0, set Y = Y

− (Y

− 1)

/2 and go to step 4.

Otherwise, go to step 1.

Step 4. Generate a random number U, and set

Z =

if U …

−Y

if U >

The random variables Z and Y generated by the foregoing algorithm are indepen-

dent, with Z being normal with mean 0 and variance 1 and Y being exponential with

rate 1. (If we want the normal random variable to have mean μ and variance σ

,we

just take μ + σ Z.)

Remarks. (a) Since c =

2e/π L 1.32, the algorithm requires a geometrically

distributed number of iterations of step 2 with mean 1.32.

(b) If we want to generate a sequence of unit normal random variables, then we can

use the exponential random variable Y obtained in step 3 as the initial exponential

needed in step 1 for the next normal to be generated. Hence, on the average, we

can simulate a unit normal by generating 1.64(= 2 * 1.32 − 1) exponentials and

computing 1.32 squares. .

Section 10.2 Simulating Continuous Random Variables 445

EXAMPLE 2d Simulating normal random variables: the polar method

It was shown in Example 7b of Chapter 6 that if X and Y are independent unit normal

random variables, then their polar coordinates R =

+ Y

, ⌰ = tan

−1

(Y/X)

are independent, with R

being exponentially distributed with mean 2 and ⌰ being

uniformly distributed on (0, 2π). Hence, if U

and U

are random numbers, then,

using the result of Example 2a, we can set

R = (−2logU

)

1/2

Θ = 2πU

from which it follows that

X = R cos Θ = (−2logU

)

1/2

cos(2πU

)

Y = R sin Θ = (−2logU

)

1/2

sin(2πU

) (2.5)

are independent unit normals. .

The preceding approach to generating unit normal random variables is called the

Box–Muller approach. Its efﬁciency suffers somewhat from its need to compute the

sine and cosine values. There is, however, a way to get around this potentially time-

consuming difﬁculty. To begin, note that if U is uniform on (0, 1) then 2U is uniform

on (0, 2), so 2U − 1 is uniform on (−1, 1). Thus, if we generate random numbers U

and U

and set

= 2U

− 1

= 2U

− 1

then (V

, V

) is uniformly distributed in the square of area 4 centered at (0, 0). (See

Figure 10.2.)

Suppose now that we continually generate such pairs (V

, V

) until we obtain one

that is contained in the disk of radius 1 centered at (0, 0)—that is, until V

+ V

… 1.

It then follows that such a pair (V

, V

) is uniformly distributed in the disk. Now, let

(1, 1)

(1, –1)

(–1, 1)

(–1, –1)

␪

+= 1

= (0, 0)

= (

)

FIGURE 10.2:

446 Chapter 10 Simulation

R, ⌰ denote the polar coordinates of this pair. Then it is easy to verify that R and ⌰

are independent, with R

being uniformly distributed on (0, 1) and ⌰ being uniformly

distributed on (0, 2π). (See Problem 13.)

Since

sin

⌰ =

+ V

cos ⌰ =

+ V

it follows from Equation (2.5) that we can generate independent unit normals X and

Y by generating another random number U and setting

X = (−2logU)

1/2

Y = (−2logU)

1/2

In fact, because (conditional on V

+ V

… 1) R

is uniform on (0, 1) and is inde-

pendent of

θ, we can use it instead of generating a new random number U, thus

showing that

X = (−2log

)

1/2

−2logS

Y = (−2logR

)

1/2

−2logS

are independent unit normals, where

S =

= V

+ V

Summing up, we have the following approach to generating a pair of independent

unit normals:

Step 1. Generate random numbers U

and U

Step 2. Set V

= 2U

− 1, V

= 2U

− 1, S = V

+ V

Step 3. If S > 1, return to step 1.

Step 4. Return the independent unit normals

X =

−2logS

, Y =

−2logS

The preceding algorithm is called the polar method. Since the probability that a

random point in the square will fall within the circle is equal to π/4 (the area of the

circle divided by the area of the square), it follows that, on average, the polar method

will require 4/π L 1.273 iterations of step 1. Hence, it will, on average, require 2.546

random numbers, 1 logarithm, 1 square root, 1 division, and 4.546 multiplications to

generate 2 independent unit normals.

EXAMPLE 2e Simulating a chi-squared random variable

The chi-squared distribution with n degrees of freedom is the distribution of χ

+ ··· + Z

, where Z

, i = 1, ..., n are independent unit normals. Now, it was

Section 10.3 Simulating from Discrete Distributions 447

shown in Section 6.3 of Chapter 6 that Z

+ Z

has an exponential distribution with

rate

. Hence, when n is even (say, n = 2k), χ

has a gamma distribution with param-

eters





. Thus, −2log(

i=1

) has a chi-squared distribution with 2k degrees

of freedom. Accordingly, can simulate a chi-squared random variable with 2k + 1

degrees of freedom by ﬁrst simulating a unit normal random variable Z and then

adding Z

to the foregoing. That is,

2k+1

= Z

− 2log

⎛

⎝



i=1

⎞

⎠

where Z, U

, ..., U

are independent, Z is a unit normal, and U

, ..., U

are uniform

(0, 1) random variables.

10.3 SIMULATING FROM DISCRETE DISTRIBUTIONS

All of the general methods for simulating random variables from continuous distribu-

tions have analogs in the discrete case. For instance, if we want to simulate a random

variable Z having probability mass function

P{X = x

}=P

, j = 0, 1, ...,



= 1

we can use the following discrete time analog of the inverse transform technique:

To simulate X for which P{X = x

}=P

, let U be uniformly distributed over (0, 1)

and set

X =

⎧

⎪

⎨

⎪

⎩

if U … P

if P

< U … P

+ P

j−1



< U …



Since

P{X = x

}=P

⎧

⎨

⎩

j−1



< U …



⎫

⎬

⎭

= P

it follows that X has the desired distribution.

EXAMPLE 3a The geometric distribution

Suppose that independent trials, each of which results in a “success” with probability

p,0 < p < 1, are continually performed until a success occurs. Letting X denote the

necessary number of trials; then

P{X = i}=(1 − p)

i−1

piÚ 1

448 Chapter 10 Simulation

which is seen by noting that X = i if the ﬁrst i − 1 trials are all failures and the ith

trial is a success. The random variable X is said to be a geometric random variable

with parameter p. Since

j−1



i=1

P{X = i}=1 − P{X > j − 1}

= 1 − P{ﬁrst j − 1 are all failures}

= 1 − (1 − p)

j−1

j Ú 1

we can simulate such a random variable by generating a random number U and then

setting X equal to that value j for which

1 − (1 − p)

j−1

< U … 1 − (1 − p)

or, equivalently, for which

(1 − p)

… 1 − U <(1 − p)

j−1

Since 1 − U has the same distribution as U, we can deﬁne X by

X = min{j : (1 − p)

… U}

= min{j : j log(1 − p) … log U}

= min

j : j Ú

log U

log(1 − p)

where the inequality has changed sign because log(1 − p) is negative [since log(1 −

p)<log 1 = 0]. Using the notation [x] for the integer part of x (that is, [x]isthe

largest integer less than or equal to x), we can write

X = 1 +



log U

log(1 − p)



As in the continuous case, special simulating techniques have been developed for

the more common discrete distributions. We now present two of these.

EXAMPLE 3b Simulating a binomial random variable

A binomial (n, p) random variable can easily be simulated by recalling that it can

be expressed as the sum of n independent Bernoulli random variables. That is, if

, ..., U

are independent uniform (0, 1) variables, then letting

1ifU

< p

0 otherwise

it follows that X K



i=1

is a binomial random variable with parameters n and p.

Section 10.4 Variance Reduction Techniques 449

EXAMPLE 3c Simulating a Poisson random variable

To simulate a Poisson random variable with mean λ, generate independent uniform

(0, 1) random variables U

, U

, ...stopping at

N = min

⎧

⎨

⎩



i=1

< e

−λ

⎫

⎬

⎭

The random variable X K N − 1 has the desired distribution. That is, if we continue

generating random numbers until their product falls below e

−λ

, then the number

required, minus 1, is Poisson with mean λ.

That X K N − 1 is indeed a Poisson random variable having mean λ can perhaps

be most easily seen by noting that

X + 1 = min

⎧

⎨

⎩



i=1

< e

−λ

⎫

⎬

⎭

is equivalent to

X = max

⎧

⎨

⎩



i=1

Ú e

−λ

⎫

⎬

⎭

where



i=1

K 1

or, taking logarithms, to

X = max

⎧

⎨

⎩



i=1

log U

Ú −λ

⎫

⎬

⎭

X = max

⎧

⎨

⎩



i=1

−log U

… λ

⎫

⎬

⎭

However, −log U

is exponential with rate 1, so X can be thought of as being the

maximum number of exponentials having rate 1 that can be summed and still be less

than λ. But by recalling that the times between successive events of a Poisson process

having rate 1 are independent exponentials with rate 1, it follows that X is equal to the

number of events by time λ of a Poisson process having rate 1; thus X has a Poisson

distribution with mean λ. .

10.4 VARIANCE REDUCTION TECHNIQUES

Let X

, ..., X

have a given joint distribution, and suppose that we are interested in

computing

θ K E[g(X

, ..., X

)]

where g is some speciﬁed function. It sometimes turns out that it is extremely difﬁcult

to analytically compute θ, and when such is the case, we can attempt to use simulation

to estimate θ . This is done as follows: Generate X

(1)

, ..., X

(1)

having the same joint

distribution as X

, ..., X

and set

= g(X

(1)

, ..., X

(1)

)

450 Chapter 10 Simulation

Now let X

(2)

, ..., X

(2)

simulate a second set of random variables (independent of the

ﬁrst set) having the distribution of X

, ..., X

and set

= g(X

(2)

, ..., X

(2)

)

Continue this until you have generated k (some predetermined number) sets and so

have also computed Y

, Y

, ..., Y

. Now, Y

, ..., Y

are independent and identically

distributed random variables, each having the same distribution as g(X

, ..., X

Thus, if we let

Y denote the average of these k random variables—that is, if

Y =



i=1

then

Y] = θ

E[(

Y − θ)

] = Var(Y)

Hence, we can use

Y as an estimate of θ . Since the expected square of the difference

between

Y and θ is equal to the variance of Y, we would like this quantity to be as

small as possible. [In the preceding situation, Var(

Y) = Var(Y

)/k, which is usually

not known in advance, but must be estimated from the generated values Y

, ..., Y

We now present three general techniques for reducing the variance of our estimator.

10.4.1 Use of Antithetic Variables

In the foregoing situation, suppose that we have generated Y

and Y

, which are

identically distributed random variables having mean θ. Now,

Var



+ Y



[Var(Y

) + Var(Y

) + 2Cov(Y

, Y

)]

Var(Y

)

Cov(Y

, Y

)

Hence, it would be advantageous (in the sense that the variance would be reduced)

if Y

and Y

were negatively correlated rather than being independent. To see how

we could arrange this, let us suppose that the random variables X

, ..., X

are inde-

pendent and, in addition, that each is simulated via the inverse transform technique.

That is, X

is simulated from F

−1

), where U

is a random number and F

is the

distribution of X

. Thus, Y

can be expressed as

= g(F

−1

), ..., F

−1

))

Now, since 1 − U is also uniform over (0, 1) whenever U is a random number (and

is negatively correlated with U), it follows that Y

deﬁned by

= g(F

−1

(1 − U

), ..., F

−1

(1 − U

))

will have the same distribution as Y

. Hence, if Y

and Y

were negatively correlated,

then generating Y

by this means would lead to a smaller variance than if it were

generated by a new set of random numbers. (In addition, there is a computational

savings because, rather than having to generate n additional random numbers, we

need only subtract each of the previous n numbers from 1.) Although we cannot, in

Section 10.4 Variance Reduction Techniques 451

general, be certain that Y

and Y

will be negatively correlated, this often turns out to

be the case, and indeed it can be proven that it will be so whenever g is a monotonic

function.

10.4.2 Variance Reduction by Conditioning

Let us start by recalling the conditional variance formula (see Section 7.5.4)

Var(Y) = E[Var(Y|Z)] + Var(E[Y|Z])

Now, suppose that we are interested in estimating E[g(X

, ..., X

)] by simulating

X = (X

, ..., X

) and then computing Y = g(X). If, for some random variable Z

we can compute E[Y|Z], then, since Var(Y|Z) Ú 0, it follows from the preceding

conditional variance formula that

Var(E[Y|Z]) … Var(Y)

Thus, since E[E[Y|Z]] = E[Y], it follows that E[Y|Z] is a better estimator of E[Y]

than is Y.

EXAMPLE 4a Estimation of π

Let U

and U

be random numbers and set V

= 2U

− 1, i = 1, 2. As noted in

Example 2d,(V

, V

) will be uniformly distributed in the square of area 4 centered at

(0, 0). The probability that this point will fall within the inscribed circle of radius 1

centered at (0, 0) (see Figure 10.2) is equal to π/4 (the ratio of the area of the circle

to that of the square). Hence, upon simulating a large number n of such pairs and

setting

1ifthejth pair falls within the circle

0 otherwise

it follows that I

, j = 1, ..., n, will be independent and identically distributed random

variables having E[I

] = π/4. Thus, by the strong law of large numbers,

+ ··· + I

→

as n

→q

Therefore, by simulating a large number of pairs (V

, V

) and multiplying the propor-

tion of them that fall within the circle by 4, we can accurately approximate π.

The preceding estimator can, however, be improved upon by using conditional

expectation. If we let I be the indicator variable for the pair (V

, V

), then, rather

than using the observed value of I, it is better to condition on V

and so utilize

E[I|V

] = P{V

+ V

… 1|V

}

= P{V

… 1 − V

}

Now,

P{V

… 1 − V

= ν}=P{V

… 1 − ν

}

= P{−

1 − ν

… V

…

1 − ν

}

1 − ν

E[I|V

] =

1 − V

452 Chapter 10 Simulation

Thus, an improvement on using the average value of I to estimate π/4istousethe

average value of

1 − V

. Indeed, since

1 − V

] =

−1

1 − ν

dν =

1 − u

du = E[

1 − U

]

where U is uniform over (0, 1), we can generate n random numbers U and use the

average value of

1 − U

as our estimate of π/4. (Problem 14 shows that this esti-

mator has the same variance as the average of the n values,

1 − V

The preceding estimator of π can be improved even further by noting that the

function g(u) =

1 − u

,0 … u … 1, is a monotonically decreasing function of u,

and so the method of antithetic variables will reduce the variance of the estimator

of E[

1 − U

]. That is, rather than generating n random numbers and using the

average value of

1 − U

as an estimator of π/4, we would obtain an improved

estimator by generating only n/2 random numbers U and then using one-half the

average of

1 − U

1 − (1 − U)

as the estimator of π/4.

The following table gives the estimates of π resulting from simulations, using n =

10, 000, based on the three estimators.

Method Estimate of π

Proportion of the random points that fall in the circle 3.1612

Average value of

1 − U

3.128448

Average value of

1 − U

1 − (1 − U)

3.139578

A further simulation using the ﬁnal approach and n = 64, 000 yielded the estimate

3.143288. .

10.4.3 Control Variates

Again, suppose that we want to use simulation to estimate E[g(X)], where X =

, ..., X

). But suppose now that, for some function f , the expected value of f (X)

is known—say, it is E[f (X)] = μ. Then, for any constant a, we can also use

W = g(X) + a[f (X) − μ]

as an estimator of E[g(X)]. Now,

Var(W) = Var[g(X)] + a

Var[f (X)] + 2a Cov[g(X), f (X)] (4.1)

Simple calculus shows that the foregoing is minimized when

a =

−Cov[f (X), g(X)]

Var[f (X)]

(4.2)

and for this value of a,

Var(W) = Var[g(X)] −

[Cov[f (X), g(X)]

Var[f (X)]

(4.3)

Problems 453

Unfortunately, neither Var[f (X)] nor Cov[f (X)], g(X)] is usually known, so we can-

not in general obtain the foregoing reduction in variance. One approach in practice

is to use the simulated data to estimate these quantities. This approach usually yields

almost all of the theoretically possible reduction in variance.

SUMMARY

Let F be a continuous distribution function and U a uniform (0, 1) random variable.

Then the random variable F

−1

(U) has distribution function F, where F

−1

(u) is that

value x such that F(x) = u. Applying this result, we can use the values of uniform (0,

1) random variables, called random numbers, to generate the values of other random

variables. This technique is called the inverse transform method.

Another technique for generating random variables is based on the rejection method.

Suppose that we have an efﬁcient procedure for generating a random variable from

the density function g and that we desire to generate a random variable having den-

sity function f . The rejection method for accomplishing this starts by determining a

constant c such that

max

f (x)

g(x)

… c

It then proceeds as follows:

1. Generate Y having density g.

2. Generate a random number U.

3. If U … f (Y)/cg(Y), set X = Y and stop.

4. Return to step 1.

The number of passes through step 1 is a geometric random variable with mean c.

Standard normal random variables can be efﬁciently simulated by the rejection

method (with g being exponential with mean 1) or by the technique known as the

polar algorithm.

To estimate a quantity θ, one often generates the values of a partial sequence

of random variables whose expected value is θ. The efﬁciency of this approach is

increased when these random variables have a small variance. Three techniques that

can often be used to specify random variables with mean θ and relatively small vari-

ances are

1. the use of antithetic variables,

2. the use of conditional expectations, and

3. the use of control variates.

PROBLEMS

10.1. The following algorithm will generate a random

permutation of the elements 1, 2, ..., n.Itissome-

what faster than the one presented in Example 1a

but is such that no position is ﬁxed until the algo-

rithm ends. In this algorithm, P(i) can be inter-

preted as the element in position i.

Step 1. Set k = 1.

Step 2. Set P (1) = 1.

Step 3. If k = n, stop. Otherwise, let k = k + 1.

Step 4. Generate a random number U and let

P(k) = P([kU] + 1)

P([kU] + 1) = k

Go to step 3.

(a) Explain in words what the algorithm is doing.

(b) Show that at iteration k—that is, when the

value of P(k) is initially set—P(1), P(2), ...,

P(k) is a random permutation of 1, 2, ..., k.

454 Chapter 10 Simulation

Hint: Use induction and argue that

, i

, ..., i

j−1

, k, i

, ..., i

k−2

, i}

= P

k−1

, i

, ..., i

j−1

, i, i

, ..., i

k−2

}

by the induction hypothesis

10.2. Develop a technique for simulating a random vari-

able having density function

f (x) =

−q < x < 0

−2x

0 < x < q

10.3. Give a technique for simulating a random variable

having the probability density function

f (x) =

⎧

⎪

⎨

⎪

⎩

(x − 2) 2 … x … 3



2 −



3 < x … 6

0 otherwise

10.4. Present a method for simulating a random variable

having distribution function

F(x) =

⎧

⎪

⎨

⎪

⎩

0 x … −3

−3 < x < 0

0 < x … 4

1 x > 4

10.5. Use the inverse transformation method to present

an approach for generating a random variable

from the Weibull distribution

F(t) = 1 − e

−at

t Ú 0

10.6. Give a method for simulating a random variable

having failure rate function

(a) λ(t) = c;

(b) λ(t) = ct;

;

(d) λ(t) = ct

10.7. Let F be the distribution function

F(x) = x

0 < x < 1

(a) Give a method for simulating a random vari-

able having distribution F that uses only a sin-

gle random number.

(b) Let U

, ..., U

be independent random num-

bers. Show that

P{max(U

, ..., U

) … x}=x

ulating a random variable having distribu-

tion F.

10.8. Suppose it is relatively easy to simulate from F

for

each i = 1, ..., n. How can we simulate from

(a) F(x) =

i=1

(x)?

(b) F(x) = 1 −

i=1

[1 − F

(x)]?

10.9. Suppose we have a method for simulating random

variables from the distributions F

and F

. Explain

how to simulate from the distribution

F(x) = pF

(x) + (1 − p)F

(x) 0 < p < 1

Give a method for simulating from

F(x) =

⎧

⎨

⎩

(1 − e

−3x

) +

x 0 < x … 1

(1 − e

−3x

) +

x > 1

10.10. In Example 2c we simulated the absolute value

of a unit normal by using the rejection procedure

on exponential random variables with rate 1. This

raises the question of whether we could obtain a

more efﬁcient algorithm by using a different expo-

nential density—that is, we could use the density

g(x) = λe

−λx

. Show that the mean number of iter-

ations needed in the rejection scheme is minimized

when λ = 1.

10.11. Use the rejection method with g(x) = 1, 0 < x < 1,

to determine an algorithm for simulating a random

variable having density function

f (x) =

60x

(1 − x)

0 < x < 1

0 otherwise

10.12. Explain how you could use random numbers to

approximate

k(x) dx,wherek(x) is an arbitrary

function.

Hint:IfU is uniform on (0, 1), what is E[k(U)]?

10.13. Let (X, Y) be uniformly distributed in the circle

of radius 1 centered at the origin. Its joint density

is thus

f (x, y) =

0 … x

+ y

… 1

Let R = (X

+ Y

)

1/2

and θ = tan

−1

(Y/X) denote

the polar coordinates of (X, Y). Show that R and

θ are independent, with R

being uniform on (0, 1)

and θ being uniform on (0, 2π).

10.14. In Example 4a, we showed that

E[(1 − V

)

1/2

] = E[(1 − U

)

1/2

] =

Reference 455

when V is uniform (−1, 1) and U is uniform (0, 1).

Now show that

Var[(1 − V

)

1/2

] = Var[(1 − U

)

1/2

]

and ﬁnd their common value.

10.15. (a) Verify that the minimum of (4.1) occurs when

a is as given by (4.2).

(b) Verify that the minimum of (4.1) is given by

(4.3).

10.16. Let X be a random variable on (0, 1) whose den-

sity is f (x). Show that we can estimate

g(x) dx

by simulating X andthentakingg(X)/f (X) as

our estimate. This method, called importance sam-

pling, tries to choose f similar in shape to g,sothat

g(X)/f (X) has a small variance.

SELF-TEST PROBLEMS AND EXERCISES

10.1. The random variable X has probability density

function

f (x) = Ce

0 < x < 1

(a) Find the value of the constant C.

(b) Give a method for simulating such a random

variable.

10.2. Give an approach for simulating a random variable

having probability density function

f (x) = 30(x

− 2x

+ x

) 0 < x < 1

10.3. Give an efﬁcient algorithm to simulate the value of

a random variable with probability mass function

= .15 p

= .2 p

= .35 p

= .30

10.4. If X is a normal random variable with mean μ

and variance σ

, deﬁne a random variable Y that

has the same distribution as X and is negatively

correlated with it.

10.5. Let X and Y be independent exponential random

variables with mean 1.

(a) Explain how we could use simulation to esti-

mate E[e

(b) Show how to improve the estimation

approach in part (a) by using a control variate.

REFERENCE

[1] ROSS,S.M.Simulation. 4th ed. San Diego: Academic Press, Inc., 2006.

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Answers to Selected Problems

CHAPTER 1

1. 67,600,000; 19,656,000 2. 1296 4. 24; 4 5. 144; 18 6. 2401 7. 720; 72; 144;

72 8. 120; 1260; 34,650 9. 27,720 10. 40,320; 10,080; 1152; 2880; 384 11. 720;

72; 144 12. 24,300,000; 17,100,720 13. 190 14. 2,598,960 16. 42; 94

17. 604,800 18. 600 19. 896; 1000; 910 20. 36; 26 21. 35 22. 18 23. 48

25. 52!/(13!)

27. 27,720 28. 65,536; 2520 29. 12,600; 945 30. 564,480

31. 165; 35 32. 1287; 14,112 33. 220; 572

CHAPTER 2

9. 74 10. .4; .1 11. 70; 2 12. .5; .32; 149/198 13. 20,000; 12,000; 11,000; 68,000;

10,000 14. 1.057 15. .0020; .4226; .0475; .0211; .00024 17. 9.10947 * 10

−6

18. .048 19. 5/18 20. .9052 22. (n + 1)/2

23. 5/12 25. .4 26. .492929

27. .0888; .2477; .1243; .2099 30. 1/18; 1/6; 1/2 31. 2/9; 1/9 33. 70/323

36. .0045; .0588 37. .0833; .5 38. 4 39. .48 40. 1/64; 21/64; 36/64; 6/64

41. .5177 44. .3; .2; .1 46. 5 48. 1.0604 * 10

−3

49. .4329 50. 2.6084 * 10

−6

52. .09145; .4268 53. 12/35 54. .0511 55. .2198; .0343

CHAPTER 3

1. 1/3

2. 1/6; 1/5; 1/4; 1/3; 1/2; 1 3. .339 5. 6/91 6. 1/2 7. 2/3

8. 1/2 9. 7/11 10. .22 11. 1/17; 1/33 12. .504; .3629 14. 35/768; 210/768

15. .4848 16. .9835 17. .0792; .264 18. .331; .383; .286; 48.62 19. 44.29;

41.18 20. .4; 1/26 21. .496; 3/14; 9/62 22. 5/9; 1/6; 5/54 23. 4/9; 1/2 24. 1/3;

1/2 26. 20/21; 40/41 28. 3/128; 29/1536 29. .0893 30. 7/12; 3/5 33. .76,

49/76 34. 27/31 35. .62, 10/19 36. 1/2 37. 1/3; 1/5; 1 38. 12/37 39. 46/185

40. 3/13; 5/13; 5/52; 15/52 41. 43/459 42. 34.48 43. 4/9 45. 1/11 48. 2/3

50. 17.5; 38/165; 17/33 51. .65; 56/65; 8/65; 1/65; 14/35; 12/35; 9/35 52. .11; 16/89;

12/27; 3/5; 9/25 55. 9 57. (c) 2/3

60. 2/3; 1/3; 3/4 61. 1/6; 3/20 65. 9/13;

1/2 69. 9; 9; 18; 110; 4; 4; 8; 120 all over 128 70. 1/9; 1/18 71. 38/64; 13/64; 13/64

73. 1/16; 1/32; 5/16; 1/4; 31/32 74. 9/19 75. 3/4, 7/12 78. p

/(1 − 2p + 2p

)

79. .5550 81. .9530 83. .5; .6; .8 84. 9/19; 6/19; 4/19; 7/15; 53/165; 7/33

89. 97/142; 15/26; 33/102

CHAPTER 4

1. p(4) = 6/91; p(2) = 8/91; p(1) = 32/91; p(0) = 1/91; p(−1) = 16/91;

p(−2) = 28/91 4. (a) 1/2; 5/18; 5/36; 5/84; 5/252; 1/252; 0; 0; 0; 0 5. n − 2i;

457

458 Answers to Selected Problems

i = 0, ..., n 6. p(3) = p(−3) = 1/8; p(1) = p(−1) = 3/8 12. p(4) = 1/16;

p(3) = 1/8; p(2) = 1/16; p(0) = 1/2; p(−i) = p(i); p(0) = 1 13. p(0) = .28;

p(500) = .27, p

(1000) = .315; p(1500) = .09; p(2000) = .045 14. p(0) = 1/2;

p(1) = 1/6; p(2) = 1/12; p(3) = 1/20; p(4) = 1/5 17. 1/4; 1/6; 1/12; 1/2 19. 1/2;

1/10; 1/5; 1/10; 1/10 20. .5918; no; −.108 21. 39.28; 37 24. p = 11/18;

maximum = 23/72 25. .46, 1.3 26. 11/2; 17/5 27. A(p + 1/10) 28. 3/5

31. p

∗

32. 11 − 10(.9)

33. 3 35. −.067; 1.089 37. 82.2; 84.5 39. 3/8

40. 11/243 42. p Ú 1/2 45. 3 50. 1/10; 1/10 51. e

−.2

;1 − 1.2e

−.2

53. 1 − e

−.6

;1 − e

−219.18

56. 253 57. .5768; .6070 59. .3935; .3033; .0902

60. .8886 61. .4082 63. .0821; .2424 65. .3935; .2293; .3935 66. 2/(2n − 1);

2/(2n − 2); e

−1

67. 2/n; (2n − 3)/(n − 1)

; e

−2

68. e

−10e

−5

70. p + (1 − p)e

−λt

71. .1500; .1012 73. 5.8125 74. 32/243; 4864/6561;

160/729; 160/729 78. 18(17)

n−1

/(35)

81. 3/10; 5/6; 75/138

82. .3439 83. 1.5

CHAPTER 5

2. 3.5e

−5/2

3. no; no 4. 1/2 5. 1 − (.01)

1/5

6. 4, 0, q 7. 3/5; 6/5 8. 2

10. 2/3; 2/3 11. 2/5 13. 2/3; 1/3 15. .7977; .6827; .3695; .9522; .1587

16. (.9938)

18. 22.66 19. 14.56 20. .9994; .75; .977 22. 9.5; .0019

23. .9258; .1762 26. .0606; .0525 28. .8363 29. .9993 32. e

−1

; e

−1/2

34. e

−1

; 1/3 38. 3/5 40. 1/y

CHAPTER 6

2. (a) 14/39; 10/39; 10/39; 5/39 (b) 84; 70; 70; 70; 40; 40; 40; 15 all divided by 429

3. 15/26; 5/26; 5/26; 1/26 4. 25/169; 40/169; 40/169; 64/169 7. p(i, j) = p

(1 − p)

i+j

8. c = 1/8; E[X] = 0 9. (12x

+ 6x)/7; 15/56; .8625; 5/7; 8/7 10. 1/2; 1 − e

−a

11. .1458 12. 39.3e

−5

13. 1/6; 1/2 15. π/4 16. n(1/2)

n−1

17. 1/3 18. 7/9

19. 1/2 21. 2/5; 2/5 22. no; 1/3 23. 1/2; 2/3; 1/20; 1/18 25. e

−1

/i! 28.

−t

;

1−3e

−2

29. .0326 30. .3772; .2061 31. .0829; .3766 32. e

−2

;1 − 3e

−2

35. 5/13; 8/13 36. 1/6; 5/6; 1/4; 3/4 41. (y + 1)

−x(y+1)

; xe

−xy

; e

−x

42. 1/2 + 3y/(4x) − y

/(4x

) 46. (1 − 2d/L)

47. .79297 48. 1 − e

−5λa

;

(1 − e

−λa

)

52. r/π 53. r 56. (a) u/(ν + 1)

CHAPTER 7

1. 52.5/12 2. 324; 199.6 3. 1/2; 1/4; 0 4. 1/6; 1/4; 1/2 5. 3/2 6. 35 7. .9; 4.9;

4.2 8. (1 − (1 − p)

)/p 10. .6; 0 11. 2(n − 1)p(1 − p)

12. (3n

− n)/(4n − 2),3n

/(4n − 2) 14. m/(1 − p) 15. 1/2 18. 4

21. .9301; 87.5755 22. 14.7 23. 147/110 26. n/(n + 1);1/(n + 1) 29.

437

; 12;

123

31. 175/6 33. 14 34. 20/19; 360/361 35. 21.2; 18.929; 49.214

36. −n/36 37. 0 38. 1/8 41. 6; 112/33 42. 100/19; 16,200/6137; 10/19;

3240/6137 45. 1/2; 0 47. 1/(n − 1) 48. 6; 7; 5.8192 49. 6.06 50. 2y

51. y

/4 53. 12 54. 8 56. N(1 − e

−10/N

) 57. 12.5 63. −96/145 65. 5.16

66. 218 67. x[1 + (2p − 1)

]

69. 1/2; 1/16; 2/81 70. 1/2, 1/3

72. 1/i; [i(i + 1)]

−1

; q 73. μ; 1 + σ

; yes; σ

79. .176; .141

Answers to Selected Problems 459

CHAPTER 8

1. Ú19/20 2. 15/17; Ú3/4; Ú10 3. Ú3 4. …4/3; .8428 5. .1416 6. .9431

7. .3085 8. .6932 9. (327)

10. 117 11. Ú.057 13. .0162; .0003;

.2514; .2514 14. n Ú 23 16. .013; .018; .691 18. ….2 23. .769; .357;

.4267; .1093; .112184

CHAPTER 9

1. 1/9; 5/9 3. .9735; .9098; .7358; .5578 10. (b) 1/6 14. 2.585; .5417; 3.1267

15. 5.5098

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Solutions to Self-Test Problems and

Exercises

CHAPTER 1

1.1. (a) There are 4! different orderings of the letters C, D, E, F. For each of these

orderings, we can obtain an ordering with A and B next to each other by

inserting A and B, either in the order A, B or in the order B, A, in any

of 5 places, namely, either before the ﬁrst letter of the permutation of

C, D, E, F, or between the ﬁrst and second, and so on. Hence, there are

2 · 5 · 4! = 240 arrangements. Another way of solving this problem is

to imagine that B is glued to the back of A. Then there are 5! orderings

in which A is immediately before B. Since there are also 5! orderings in

which B is immediately before A, we again obtain a total of 2 · 5! = 240

different arrangements.

(b) There are 6! = 720 possible arrangements, and since there are as many

with A before B as with B before A, there are 360 arrangements.

B before C as have any of the 3! possible orderings of A, B, and C. Hence,

there are 720/6 = 120 possible orderings.

(d) Of the 360 arrangements that have A before B, half will have C before D

and half D before C. Hence, there are 180 arrangements having A before

B and C before D.

(e) Gluing B to the back of A and D to the back of C yields 4! = 24 different

orderings in which B immediately follows A and D immediately follows

C. Since the order of A and B and of C and D can be reversed, there are

4 · 24 = 96 different arrangements.

(f) There are 5! orderings in which E is last. Hence, there are 6! − 5! = 600

orderings in which E is not last.

1.2. 3! 4! 3! 3!, since there are 3! possible orderings of countries and then the coun-

trymen must be ordered.

1.3. (a) 10 · 9 · 8 = 720

(b) 8 · 7 · 6 + 2 · 3 · 8 · 7 = 672. The result of part (b) follows because

there are 8 · 7 · 6 choices not including A or B and there are 3 · 8 · 7

choices in which a speciﬁed one of A and B, but not the other, serves. The

latter follows because the serving member of the pair can be assigned to

any of the 3 ofﬁces, the next position can then be ﬁlled by any of the other

8 people, and the ﬁnal position by any of the remaining 7.

6 + 3 · 2 · 8 = 384.

(d) 3 · 9 · 8 = 216.

(e) 9 · 8 · 7 + 9 · 8 = 576.

461

462 Solutions to Self-Test Problems and Exercises

1.4. (a)





(b)



















1.5.



3, 2, 2



= 210

1.6. There are





= 35 choices of the three places for the letters. For each choice,

there are (26)

(10)

different license plates. Hence, altogether there are 35 ·

(26)

· (10)

different plates.

1.7. Any choice of r of the n items is equivalent to a choice of n − r, namely, those

items not selected.

1.8. (a) 10 · 9 · 9 ···9 = 10 · 9

n−1

(b)





n−i

, since there are





choices of the i places to put the zeroes and

then each of the other n − i positions can be any of the digits 1, ...,9.

1.9. (a)





(b) 3





(c)







= 3n

(n − 1)

(d) n

(e)





= 3





+ 3n

(n − 1) + n

1.10. There are 9 · 8 · 7 · 6 · 5 numbers in which no digit is repeated. There are





· 8 · 7 · 6 numbers in which only one speciﬁed digit appears twice, so

there are 9





· 8 · 7 · 6 numbers in which only a single digit appears twice.

There are 7 ·

2!2!

numbers in which two speciﬁed digits appear twice, so there

are





7 ·

2!2!

numbers in which two digits appear twice. Thus, the answer is

9 · 8 · 7 · 6 · 5 + 9





· 8 · 7 · 6 +





7 ·

2!2!

1.11. (a) We can regard this as a seven-stage experiment. First choose the 6 mar-

ried couples that have a representative in the group, and then select one of

the members of each of these couples. By the generalized basic principle

of counting, there are





different choices.

(b) First select the 6 married couples that have a representative in the group,

and then select the 3 of those couples that are to contribute a man. Hence,

there are







10!

4!3!3!

different choices. Another way to solve this is

to ﬁrst select 3 men and then select 3 women not related to the selected

men. This shows that there are







10!

3!3!4!

different choices.

Solutions to Self-Test Problems and Exercises 463

1.12.













= 3430. The ﬁrst term gives the number of committees

that have 3 women and 3 men; the second gives the number that have 4 women

and 2 men.

1.13. (number of solutions of x

+···+x

= 4)(number of solutions of x

+···+

= 5)(number of solutions of x

+ ··· + x

= 6) =







1.14. Since there are



j − 1

n − 1



positive vectors whose sum is j, there must be



j=n



j − 1

n − 1



such vectors. But



j − 1

n − 1



is the number of subsets of size n

from the set of numbers {1, ..., k} in which j is the largest element in the sub-

set. Consequently,



j=n



j − 1

n − 1



is just the total number of subsets of size n

from a set of size k, showing that the preceding answer is equal to





1.15. Let us ﬁrst determine the number of different results in which k people pass.

Because there are





different groups of size k and k! possible orderings of

their scores, it follows that there are





k! possible results in which k people

pass. Consequently, there are



k=0





k! possible results.

1.16. The number of subsets of size 4 is





= 4845. Because the number of these

that contain none of the ﬁrst ﬁve elements is





= 1365, the number that

contain at least one is 3480. Another way to solve this problem is to note that

there are





4−i



that contain exactly i of the ﬁrst ﬁve elements and sum this

for i = 1, 2, 3, 4.

1.17. Multiplying both sides by 2, we must show that

n(n − 1) = k(k − 1) + 2k(n − k) + (n − k)(n − k − 1)

This follows because the right side is equal to

(1 − 2 + 1) + k(−1 + 2n − n − n + 1) + n(n − 1)

For a combinatorial argument, consider a group of n items and a subgroup

of k of the n items. Then





is the number of subsets of size 2 that contain

2 items from the subgroup of size k, k(n − k) is the number that contain 1

item from the subgroup, and



n−k



is the number that contain 0 items from the

subgroup. Adding these terms gives the total number of subgroups of size 2,

namely,





1.18. There are 3 choices that can be made from families consisting of a single parent

and 1 child; there are 3 · 1 · 2 = 6 choices that can be made from families

consisting of a single parent and 2 children; there are 5 · 2 · 1 = 10 choices

that can be made from families consisting of 2 parents and a single child; there

are 7 · 2 · 2 = 28 choices that can be made from families consisting of 2 parents

and 2 children; there are 6 · 2 · 3 = 36 choices that can be made from families

consisting of 2 parents and 3 children. Hence, there are 80 possible choices.

464 Solutions to Self-Test Problems and Exercises

1.19. First choose the 3 positions for the digits, and then put in the letters and digits.

Thus, there are





· 26 · 25 · 24 · 23 · 22 · 10 · 9 · 8 different plates. If the

digits must be consecutive, then there are 6 possible positions for the digits,

showing that there are now 6 · 26 · 25 · 24 · 23 · 22 · 10 · 9 · 8 different

plates.

CHAPTER 2

2.1. (a) 2 · 3 · 4 = 24

(b) 2 · 3 = 6

(d) AB ={(c, pasta, i), (c, rice, i), (c, potatoes, i)}

(e) 8

(f) ABC ={(c, rice, i)}

2.2. Let A be the event that a suit is purchased, B be the event that a shirt is pur-

chased, and C be the event that a tie is purchased. Then

P(A

∪ B ∪ C) = .22 + .30 + .28 − .11 − .14 − .10 + .06 = .51

(a) 1 − .51 = .49

(b) The probability that two or more items are purchased is

P(AB ∪ AC ∪ BC) = .11 + .14 + .10 − .06 − .06 − .06 + .06 = .23

Hence, the probability that exactly 1 item is purchased is .51 − .23 = .28.

2.3. By symmetry, the 14th card is equally likely to be any of the 52 cards; thus, the

probability is 4/52. A more formal argument is to count the number of the 52!

outcomes for which the 14th card is an ace. This yields

p =

4 · 51 · 50 ···2 · 1

(52)!

Letting A be the event that the ﬁrst ace occurs on the 14th card, we have

P(A) =

48 · 47 ···36 · 4

52 · 51 ···40 · 39

= .0312

2.4. Let D denote the event that the minimum temperature is 70 degrees. Then

P(A ∪ B) = P(A) + P(B) − P(AB) = .7 − P(AB)

P(C ∪ D) = P(C ) + P(D) − P(CD) = .2 + P(D) − P(DC)

Since A ∪ B = C ∪ D and AB = CD, subtracting one of the preceding

equations from the other yields

0 = .5 − P(D)

P(D) = .5.

2.5. (a)

52 · 48 · 44 · 40

52 · 51 · 50 · 49

= .6761

(b)

52 · 39 · 26 · 13

52 · 51 · 50 · 49

= .1055

Solutions to Self-Test Problems and Exercises 465

2.6. Let R be the event that both balls are red, and let B be the event that both are

black. Then

P(R ∪ B) = P(R) + P(B) =

3 · 4

6 · 10

3 · 6

6 · 10

= 1/2

2.7. (a)

⎛

⎝

⎞

⎠

= 1.3 * 10

−8

(b)

⎛

⎝

⎞

⎠

⎛

⎝

⎞

⎠

⎛

⎝

⎞

⎠

= 3.3 * 10

−6

(c)

⎛

⎝

⎞

⎠

⎛

⎝

⎞

⎠

⎛

⎝

⎞

⎠

+ 1.3 * 10

−8

+ 3.3 * 10

−6

= 1.8 * 10

−4

2.8. (a)

3 · 4 · 4 · 3





= .1439

(b)

⎛

⎝

⎞

⎠

⎛

⎝

⎞

⎠

⎛

⎝

⎞

⎠

= .0360

(c)

⎛

⎝

⎞

⎠

⎛

⎝

⎞

⎠

= .0699

2.9. Let S =



i=1

, and consider the experiment of randomly choosing an element

of S. Then P(A) = N(A)/N(S), and the results follow from Propositions 4.3

and 4.4.

2.10. Since there are 5! = 120 outcomes in which the position of horse number

1 is speciﬁed, it follows that N(A) = 360. Similarly, N(B) = 120, and

N(AB) = 2 · 4! = 48. Hence, from Self-Test Problem 9, we obtain

N(A ∪ B) = 432.

2.11. One way to solve this problem is to start with the complementary probability

that at least one suit does not appear. Let A

, i = 1, 2, 3, 4, be the event that no

cards from suit i appear. Then

⎛

⎝



i=1

⎞

⎠



P(A

) −



i:i<j

P(A

) + ··· − P(A

)

466 Solutions to Self-Test Problems and Exercises

= 4









−

























= 4









− 6









+ 4









The desired probability is then 1 minus the preceding. Another way to solve is

to let A be the event that all 4 suits are represented, and then use

P(A) = P(n, n, n, n, o) + P(n, n, n, o, n) + P(n, n, o, n, n) + P(n, o, n, n, n)

where P (n, n, n, o, n), for instance, is the probability that the ﬁrst card is from a

new suit, the second is from a new suit, the third is from a new suit, the fourth

is from an old suit (that is, one which has already appeared) and the ﬁfth is

from a new suit. This gives

P(A) =

52 · 39 · 26 · 13 · 48 + 52 · 39 · 26 · 36 · 13

52 · 51 · 50 · 49 · 48

52 · 39 · 24 · 26 · 13 + 52 · 12 · 39 · 26 · 13

52 · 51 · 50 · 49 · 48

52 · 39 · 26 · 13(48 + 36 + 24 + 12)

52 · 51 · 50 · 49 · 48

= .2637

2.12. There are (10)!/2

different divisions of the 10 players into a ﬁrst roommate

pair, a second roommate pair, and so on. Hence, there are (10)!/(5!2

) divi-

sions into 5 roommate pairs. There are







ways of choosing the front-

court and backcourt players to be in the mixed roommate pairs and then

2 ways of pairing them up. As there is then 1 way to pair up the

remaining two backcourt players and 4!/(2!2

) = 3 ways of making two

roommate pairs from the remaining four frontcourt players, the desired

probability is

P{2 mixed pairs}=







(2)(3)

(10)!/(5!2

)

= .5714

2.13. Let R denote the event that letter R is repeated; similarly, deﬁne the events E

and V. Then

P{same letter}=P(R) + P(E) + P(V) =

Solutions to Self-Test Problems and Exercises 467

2.14. Let B

= A

, B

= A

⎛

⎝

i−1



j=1

⎞

⎠

, i > 1. Then

⎛

⎝



i=1

⎞

⎠

= P

⎛

⎝



i=1

⎞

⎠



i=1

P(B

)

…



i=1

P(A

)

where the ﬁnal equality uses the fact that the B

are mutually exclusive. The

inequality then follows, since B

( A

2.15.

⎛

⎝



i=1

⎞

⎠

= 1 − P

⎛

⎜

⎝

⎛

⎝



i=1

⎞

⎠

⎞

⎟

⎠

= 1 − P

⎛

⎝



i=1

⎞

⎠

Ú 1 −



i=1

P(A

)

= 1

2.16. The number of partitions for which {1} is a subset is equal to the number of

partitions of the remaining n − 1 elements into k − 1 nonempty subsets,

namely, T

k−1

(n − 1). Because there are T

(n − 1) partitions of {2, ..., n − 1}

into k nonempty subsets and then a choice of k of them in which to place

element 1, it follows that there are kT

(n − 1) partitions for which {1} is not a

subset. Hence, the result follows.

2.17. Let R, W, B denote, respectively, the events that there are no red, no white,

and no blue balls chosen. Then

P(R ∪ W ∪ B) = P(R) + P(W) + P (B) − P(RW) − P(RB)

− P(WB) + P(RWB)

























−









−









−









L 0.2933

468 Solutions to Self-Test Problems and Exercises

Thus, the probability that all colors appear in the chosen subset is approxi-

mately 1 − 0.2933 = 0.7067.

2.18. (a)

8·7·6·5·4

17·16·15·14·13

221

(b) Because there are 9 nonblue balls, the probability is

9·8·7·6·5

17·16·15·14·13

442

sibilities for the ﬁnal 3 balls are equally likely, the probability is

3!·4·8·5

17·16·15

(d) The probability that the red balls are in a speciﬁed 4 spots is

4·3·2·1

17·16·15·14

Because there are 14 possible locations of the red balls where they are all

together, the probability is

14·4·3·2·1

17·16·15·14

170

2.19. (a) The probability that the 10 cards consist of 4 spades, 3 hearts, 2 diamonds,

and 1 club is











. Because there are 4! possible choices

of the suits to have 4, 3, 2, and 1 cards, respectively, it follows that the

probability is











(b) Because there are





= 6 choices of the two suits that are to have 3

cards and then 2 choices for the suit to have 4 cards, the probability is











2.20. All the red balls are removed before all the blue ones if and only if the very

last ball removed is blue. Because all 30 balls are equally likely to be the last

ball removed, the probability is 10/30.

CHAPTER 3

3.1. (a) P(no aces) =











(b) 1 − P(no aces) −









⎛

⎝

⎞

⎠

⎛

⎝

13 − i

⎞

⎠

⎛

⎝

⎞

⎠

3.2. Let L

denote the event that the life of the battery is greater than 10, 000 *

i miles.

(a) P(L

) = P(L

)/P(L

) = P(L

)/P(L

) = 1/2

(b) P(L

) = P(L

)/P(L

) = P(L

)/P(L

) = 1/8

3.3. Put 1 white and 0 black balls in urn one, and the remaining 9 white and 10

black balls in urn two.

Solutions to Self-Test Problems and Exercises 469

3.4. Let T be the event that the transferred ball is white, and let W be the event

that a white ball is drawn from urn B. Then

P(T|W) =

P(W|T)P(T)

P(W|T)P(T) + P(W|T

)P(T

)

(2/7)(2/3)

(2/7)(2/3) + (1/7)(1/3)

= 4/5

3.5. (a)

r+w

, because each of the r + w balls is equally likely to be the ith ball

removed.

(b), (c)

P(R

) =

P(R

)

P(R

)

(

)

(

r+w

)

r+w

r − 1

r + w − 1

A simpler argument is to note that, for i Z j, given that the ith removal

is a red ball, the jth removal is equally likely to be any of the remaining

r + w − 1 balls, of which r − 1 are red.

3.6. Let B

denote the event that ball i is black, and let R

= B

. Then

P(B

) =

P(R

)P(B

)

P(R

)P(B

) + P(R

)P(R

)

[r/[(b + r + c)][b/(b + r)]

[r/(b + r + c)][b/(b + r)] + [(r + c)/(b + r + c)][r/(b + r)]

b + r + c

3.7. Let B denote the event that both cards are aces.

(a)

P{B|yes to ace of spades}=

P{B, yes to ace of spades}

P{yes to ace of spades}























= 3/51

(b) Since the second card is equally likely to be any of the remaining 51, of

which 3 are aces, we see that the answer in this situation is also 3/51.

which is considered second, the result should be the same as in part (b).

A more formal argument is as follows:

470 Solutions to Self-Test Problems and Exercises

P{B|second is ace}=

P{B, second is ace}

P{second is ace}

P(B)

P(B) + P{ﬁrst is not ace, second is ace}

(4/52)(3/51)

(4/52)(3/51) + (48/52)(4/51)

= 3/51

(d)

P{B|at least one}=

P(B)

P{at least one}

(4/52)(3/51)

1 − (48/52)(47/51)

= 1/33

3.8.

P(H|E)

P(G|E)

P(HE)

P(GE)

P(H)P(E|H)

P(G)P(E|G)

Hypothesis H is 1.5 times as likely.

3.9. Let A denote the event that the plant is alive and let W be the event that it was

watered.

(a)

P(A) = P(A|W)P(W) + P(A|W

)P(W

)

= (.85)(.9) + (.2)(.1) = .785

(b)

P(W

) =

P(A

)P(W

)

P(A

)

(.8)(.1)

.215

3.10. (a) 1 − P (no red balls) = 1 −









(b) Given that no red balls are chosen, the 6 chosen are equally likely to be

any of the 22 nonred balls. Thus,

P(2 green|no red) =











3.11. Let W be the event that the battery works, and let C and D denote the events

that the battery is a type C andthatitisatypeD battery, respectively.

(a) P(W) = P(W|C)P(C) + P(W|D)P(D) = .7(8/14) + .4(6/14) = 4/7

(b)

P(C|W

) =

P(CW

)

P(W

)

P(W

|C)P(C)

3/7

.3(8/14)

3/7

= .4

Solutions to Self-Test Problems and Exercises 471

3.12. Let L

be the event that Maria likes book i, i = 1, 2. Then

P(L

) =

P(L

)

P(L

)

P(L

)

Using that L

is the union of the mutually exclusive events L

and L

,we

see that

.5 = P(L

) = P(L

) + P (L

) = .4 + P(L

)

Thus,





= .25

3.13. (a) This is the probability that the last ball removed is blue. Because each of

the 30 balls is equally likely to be the last one removed, the probability is

1/3.

(b) This is the probability that the last red or blue ball to be removed is a blue

ball. Because it is equally likely to be any of the 30 red or blue balls, the

probability that it is blue is 1/3.

, R

, G

denote, respectively, the events that the ﬁrst color removed

is blue, the second is red, and the third is green. Then

P(B

) = P(G

)P(R

)P(B

) =

where P(G

) is just the probability that the very last ball is green and

P(R

) is computed by noting that, given that the last ball is green,

each of the 20 red and 10 blue balls is equally likely to be the last of that

group to be removed, so the probability that it is one of the red balls is

20/30. (Of course, P(B

) = 1.)

(d) P(B

) = P(B

) + P (B

) =

171

3.14. Let H be the event that the coin lands heads, let T

be the event that B is told

that the coin landed heads, let F be the event that A forgets the result of the

toss, and let C be the event that B is told the correct result. Then

(a)

P(T

) = P(T

|F)P(F) + P(T

)P(F

)

= (.5)(.4) + P(H)(.6)

= .68

(b)

P(C) = P(C |F)P(F) + P(C|F

)P(F

)

= (.5)(.4) + 1(.6) = .80

(c)

P(H|T

) =

P(HT

)

P(T

)

472 Solutions to Self-Test Problems and Exercises

Now,

P(HT

) = P(HT

|F)P(F) + P(HT

)P(F

)

= P(H|F)P(T

|HF)P(F) + P(H)P(F

)

= (.8)(.5)(.4) + (.8)(.6) = .64

giving the result P(H|T

) = .64/.68 = 16/17.

3.15. Since the black rat has a brown sibling, we can conclude that both of its parents

have one black and one brown gene.

(a)

P(2 black|at least one) =

P(2)

P(at least one)

1/4

3/4

(b) Let F be the event that all 5 offspring are black, let B

be the event that

the black rat has 2 black genes, and let B

be the event that it has 1 black

and 1 brown gene. Then

P(B

|F) =

P(F|B

)P(B

)

P(F|B

)P(B

) + P(F|B

)P(B

)

(1)(1/3)

(1)(1/3) + (1/2)

(2/3)

3.16. Let F be the event that a current ﬂows from A to B, and let C

be the event

that relay i closes. Then

P(F) = P(F|C

+ P(F|C

)(1 − p

)

Now,

P(F|C

) = P(C

∪ C

)

= P(C

) + P(C

) − P (C

)

= p

+ p

− p

Also,

P(F|C

) = P(C

∪ C

)

= p

+ p

− p

Hence, for part (a), we obtain

P(F) = p

+ p

− p

) + (1 − p

+ p

− p

)

For part (b), let q

= 1 − p

. Then

P(C

|F) = P(F|C

)P(C

)/P(F)

= p

[1 − P (C

∪ C

)]/P(F)

= p

(1 − q

− q

+ q

)/P(F)

3.17. Let A be the event that component 1 is working, and let F be the event that

the system functions.

(a)

P(A|F) =

P(AF)

P(F)

P(A)

P(F)

1/2

1 − (1/2)

Solutions to Self-Test Problems and Exercises 473

where P(F) was computed by noting that it is equal to 1 minus the prob-

ability that components 1 and 2 are both failed.

(b)

P(A|F) =

P(AF)

P(F)

P(F|A)P(A)

P(F)

(3/4)(1/2)

(1/2)

+ 3(1/2)

where P(F) was computed by noting that it is equal to the probability that

all 3 components work plus the three probabilities relating to exactly 2 of

the components working.

3.18. If we assume that the outcomes of the successive spins are independent, then

the conditional probability of the next outcome is unchanged by the result that

the previous 10 spins landed on black.

3.19. Condition on the outcome of the initial tosses:

P(A odd) = P

(1 − P

)(1 − P

) + (1 − P

+ P

(A odd)

+ (1 − P

)(1 − P

)P(A odd)

so,

P(A odd) =

(1 − P

)(1 − P

) + (1 − P

+ P

− P

3.20. Let A and B be the events that the ﬁrst trial is larger and that the second is

larger, respectively. Also, let E be the event that the results of the trials are

equal. Then

1 = P(A) + P(B) + P(E)

But, by symmetry, P(A) = P(B): thus,

P(B) =

1 − P(E)

1 −



i=1

Another way of solving the problem is to note that

P(B) =



j>i

P{ﬁrst trial results in i, second trial results in j}



j>i

To see that the two expressions derived for P(B) are equal, observe that

1 =



i=1



j=1



jZi



+ 2



j>i

474 Solutions to Self-Test Problems and Exercises

3.21. Let E ={A gets more heads than B};then

P(E) = P(E|A leads after both ﬂip n)P(A leads after both ﬂip n)

+ P(E| even after both ﬂip n)P( even after both ﬂip n)

+ P(E|B leads after both ﬂip n)P(B leads after both ﬂip n)

= P(A leads) +

P( even)

Now, by symmetry,

P(A leads) = P(B leads)

1 − P(even)

Hence,

P(E) =

3.22. (a) Not true: In rolling 2 dice, let E ={sumis7}, F ={1st die does not land on 4},

and G={2nd die does not land on 3}. Then

P(E|F ∪ G) =

P{7, not (4, 3)}

P{not (4, 3)}

5/36

35/36

= 5/35 Z P(E)

(b)

P(E(F ∪ G)) = P(EF ∪ EG)

= P(EF) + P(EG) since EFG =∅

= P(E)[P(F) + P(G)]

= P(E)P(F ∪ G) since FG =∅

(c)

P(G|EF) =

P(EFG)

P(EF)

P(E)P(FG)

P(EF)

since E is independent of FG

P(E)P(F)P(G)

P(E)P(F)

by independence

= P(G).

3.23. (a) necessarily false; if they were mutually exclusive, then we would have

0 = P(AB) Z P(A)P(B)

(b) necessarily false; if they were independent, then we would have

P(AB) = P(A)P(B)>0

P(A ∪ B) = P(A) + P(B) = 1.2

(d) possibly true

Solutions to Self-Test Problems and Exercises 475

3.24. The probabilities in parts (a), (b), and (c) are .5, (.8)

= .512, and (.9)

L .4783,

respectively.

3.25. Let D

, i = 1, 2, denote the event that radio i is defective. Also, let A and B

be the events that the radios were produced at factory A and at factory B,

respectively. Then

P(D

) =

P(D

)

P(D

)

P(D

|A)P(A) + P(D

|B)P(B)

P(D

|A)P(A) + P(D

|B)P(B)

(.05)

(1/2) + (.01)

(1/2)

(.05)(1/2) + (.01)(1/2)

= 13/300

3.26. We are given that P(AB) = P(B) and must show that this implies that P(B

) =

P(A

). One way is as follows:

P(B

) = P((A ∪ B)

)

= 1 − P(A ∪ B)

= 1 − P(A) − P(B) + P(AB)

= 1 − P(A)

= P(A

)

3.27. The result is true for n = 0. With A

denoting the event that there are i red

balls in the urn after stage n, assume that

P(A

) =

n + 1

, i = 1, ..., n + 1

Now let B

, j = 1, ..., n + 2, denote the event that there are j red balls in the

urnafterstagen + 1. Then

P(B

) =

n+1



i=1

P(B

)P(A

)

n + 1

n+1



i=1

P(B

)

n + 1

[P(B

j−1

) + P (B

)]

Because there are n + 2 balls in the urn after stage n, it follows that P(B

j−1

)

is the probability that a red ball is chosen when j − 1ofthen + 2 balls in the

urn are red and P(B

) is the probability that a red ball is not chosen when j

of the n + 2 balls in the urn are red. Consequently,

P(B

j−1

) =

j − 1

n + 2

, P(B

) =

n + 2 − j

n + 2

476 Solutions to Self-Test Problems and Exercises

Substituting these results into the equation for P(B

) gives

P(B

) =

n + 1



j − 1

n + 2

n + 2 − j

n + 2



n + 2

This completes the induction proof.

3.28. If A

is the event that player i receives an ace, then

P(A

) = 1 −



2n − 2







= 1 −

n − 1

2n − 1

3n − 1

4n − 2

By arbitrarily numbering the aces and noting that the player who does not

receive ace number one will receive n of the remaining 2n − 1 cards, we

see that

P(A

) =

2n − 1

Therefore,

P(A

) = 1 − P(A

) = 1 −

P(A

)

P(A

)

n − 1

3n − 1

We may regard the card division outcome as the result of two trials, where trial

i, i = 1, 2, is said to be a success if ace number i goes to the ﬁrst player. Because

the locations of the two aces become independent as n goes to inﬁnity, with

each one being equally likely to be given to either player, it follows that the

trials become independent, each being a success with probability 1/2. Hence,

in the limiting case where n

→q, the problem becomes one of determining

the conditional probability that two heads result, given that at least one does,

when two fair coins are ﬂipped. Because

n−1

3n−1

converges to 1/3, the answer

agrees with that of Example 2b.

3.29. (a) For any permutation i

, ..., i

of 1, 2, ..., n, the probability that the suc-

cessive types collected is i

, ..., i

is p

···p

i=1

. Consequently,

the desired probability is n!

i=1

(b) For i

, ..., i

all distinct,

P(E

···E

) =



n − k



which follows because there are no coupons of types i

, ..., i

when each

of the n independent selections is one of the other n − k types. It now

follows by the inclusion–exclusion identity that

P(∪

i=1

) =



k=1

(−1)

k+1





n − k



Solutions to Self-Test Problems and Exercises 477

Because 1 − P(∪

i=1

) is the probability that one of each type is obtained,

by part (a) it is equal to

. Substituting this into the preceding equation

gives

1 −



k=1

(−1)

k+1





n − k



n! = n

−



k=1

(−1)

k+1





(n − k)

n! =



k=0

(−1)





(n − k)

3.30.

P(E|E ∪ F) = P(E|F(E ∪ F))P(F|E ∪ F) + P(E |F

(E ∪ F))P(F

|E ∪ F)

Using

F(E ∪ F) = F and F

(E ∪ F) = F

gives

P(E|E ∪ F) = P(E|F)P(F|E ∪ F) + P(E|EF

)P(F

|E ∪ F)

= P(E|F)P(F|E ∪ F) + P(F

|E ∪ F)

Ú P(E|F)P(F|E ∪ F) + P(E|F)P (F

|E ∪ F)

= P(E|F)

CHAPTER 4

4.1. Since the probabilities sum to 1, we must have 4P{X = 3}+.5 = 1, implying

that P{X = 0}=.375, P{X = 3}=.125. Hence, E[X] = 1(.3) + 2(.2) +

3(.125) = 1.075.

4.2. The relationship implies that p

= c

, i = 1, 2, where p

= P{X = i}. Because

these probabilities sum to 1, it follows that

(1 + c + c

) = 1 * p

1 + c + c

Hence,

E[X] = p

+ 2p

c + 2c

1 + c + c

4.3. Let X be the number of ﬂips. Then the probability mass function of X is

= p

+ (1 − p)

, p

= 1 − p

= 2p(1 − p)

478 Solutions to Self-Test Problems and Exercises

Hence,

E[X] = 2p

+ 3p

= 2p

+ 3(1 − p

) = 3 − p

− (1 − p)

4.4. The probability that a randomly chosen family will have i children is n

/m.

Thus,

E[X] =



i=1

Also, since there are in

children in families having i children, it follows that

the probability that a randomly chosen child is from a family with i children is



i=1

. Therefore,

E[Y] =



i=1



i=1

Thus, we must show that



i=1



i=1



i=1



i=1

or, equivalently, that



j=1



i=1



i=1



j=1

or, equivalently, that



i=1



j=1



i=1



j=1

ijn

But, for a ﬁxed pair i, j, the coefﬁcient of n

in the left-side summation of

the preceding inequality is i

+ j

, whereas its coefﬁcient in the right-hand

summation is 2ij. Hence, it sufﬁces to show that

+ j

Ú 2ij

which follows because (i − j)

Ú 0.

4.5. Let p = P{X = 1}. Then E[X] = p and Var(X) = p(1 − p),so

p = 3p(1 − p)

implying that p = 2/3. Hence, P{X = 0}=1/3.

Solutions to Self-Test Problems and Exercises 479

4.6. If you wager x on a bet that wins the amount wagered with probability p and

loses that amount with probability 1 − p, then your expected winnings are

xp − x(1 − p) = (2p − 1)x

which is positive (and increasing in x) if and only if p > 1/2. Thus, if p … 1/2,

one maximizes one’s expected return by wagering 0, and if p > 1/2, one maxi-

mizes one’s expected return by wagering the maximal possible bet. Therefore,

if the information is that the .6 coin was chosen, then you should bet 10, and if

the information is that the .3 coin was chosen, then you should bet 0. Hence,

your expected payoff is

(1.2 − 1)10 +

0 − C = 1 − C

Since your expected payoff is 0 without the information (because in this case

the probability of winning is

(.6) +

(.3)<1/2), it follows that if the infor-

mation costs less than 1, then it pays to purchase it.

4.7. (a) If you turn over the red paper and observe the value x, then your expected

return if you switch to the blue paper is

2x(1/2) + x/2(1/2) = 5x/4 > x

Thus, it would always be better to switch.

(b) Suppose the philanthropist writes the amount x on the red paper. Then

the amount on the blue paper is either 2x or x/2. Note that if x/2 Ú y,then

the amount on the blue paper will be at least y and will thus be accepted.

Hence, in this case, the reward is equally likely to be either 2x or x/2, so

E[R

(x)] = 5x/4, if x/2 Ú y

If x/2 < y … 2x, then the blue paper will be accepted if its value is 2x and

rejected if it is x/2. Therefore,

E[R

(x)] = 2x(1/2) + x(1/2) = 3x/2, if x/2 < y … 2x

Finally, if 2x < y, then the blue paper will be rejected. Hence, in this case,

the reward is x,so

(x) = x,if2x < y

That is, we have shown that when the amount x is written on the red

paper, the expected return under the y-policy is

E[R

(x)] =

⎧

⎨

⎩

x if x < y/2

3x/2ify/2 … x < 2y

5x/4ifx Ú 2y

4.8. Suppose that n independent trials, each of which results in a success with prob-

ability p, are performed. Then the number of successes will be less than or

equal to i if and only if the number of failures is greater than or equal to n − i.

But since each trial is a failure with probability 1 − p, it follows that the num-

ber of failures is a binomial random variable with parameters n and 1 − p.

Hence,

480 Solutions to Self-Test Problems and Exercises

P{Bin(n, p) … i}=P{Bin (n,1 − p) Ú n − i}

= 1 − P{Bin (n,1 − p) … n − i − 1}

The ﬁnal equality follows from the fact that the probability that the number

of failures is greater than or equal to n − i is 1 minus the probability that it is

less than n − i.

4.9. Since E[X] = np,Var(X) = np(1 − p), we are given that np = 6, np(1 − p) =

2.4. Thus, 1 − p = .4, or p = .6, n = 10. Hence,

{X = 5}=





(.6)

(.4)

4.10. Let X

, i = 1, ..., m, denote the number on the ith ball drawn. Then

P{X … k}=P{X

… k, X

… k, ..., X

… k}

= P{X

… k}P{X

… k}···P{X

… k}





Therefore,

P{X = k}=P{X … k}−P{X … k − 1}=





−



k − 1



4.11. (a) Given that A wins the ﬁrst game, it will win the series if, from then on, it

wins 2 games before team B wins 3 games. Thus,

P{A wins|A wins ﬁrst}=



i=2





(1 − p)

4−i

(b)

P{A wins ﬁrst|A wins}=

P{A wins|A wins ﬁrst}P{A wins ﬁrst}

P{A wins}



i=2





i+1

(1 − p)

4−i



i=3





(1 − p)

5−i

4.12. To obtain the solution, condition on whether the team wins this weekend:



i=3





(.4)

(.6)

4−i

+ .5



i=3





(.7)

(.3)

4−i

Solutions to Self-Test Problems and Exercises 481

4.13. Let C be the event that the jury makes the correct decision, and let F be the

event that four of the judges agreed. Then

P(C) =



i=4





(.7)

(.3)

7−i

Also,

P(C|F) =

P(CF)

P(F)





(.7)

(.3)





(.7)

(.3)





(.7)

(.3)

= .7

4.14. Assuming that the number of hurricanes can be approximated by a Poisson

random variable, we obtain the solution



i=0

−5.2

(5.2)

/i!

4.15.

E[Y] =



i=1

iP{X = i}/P{X > 0}

= E[X]/P{X > 0}

1 − e

−λ

4.16. (a) 1/n

(b) Let D be the event that girl i and girl j choose different boys. Then

P(G

) = P(G

|D)P(D) + P(G

)P(D

)

= (1/n)

(1 − 1/n)

n − 1

Therefore,

P(G

) =

n − 1

(c), (d) Because, when n is large, P (G

) is small and nearly equal to P(G

it follows from the Poisson paradigm that the number of couples is

approximately Poisson distributed with mean



i=1

P(G

) = 1. Hence,

L e

−1

and P

L e

−1

/k!

(e) To determine the probability that a given set of k girls all are coupled,

condition on whether or not D occurs, where D is the event that they all

choose different boys. This gives

482 Solutions to Self-Test Problems and Exercises

P(G

···G

) = P(G

···G

|D)P(D) + P(G

···G

)P(D

)

= P(G

···G

|D)P(D)

= (1/n)

n(n − 1) ···(n − k + 1)

(n − k)!n

Therefore,



<...<i

P(G

···G

) =





P(G

···G

) =

n!n!

(n − k)!(n − k)!k!n

and the inclusion–exclusion identity yields

1 − P

= P(∪

i=1

) =



k=1

(−1)

k+1

n!n!

(n − k)!(n − k)!k!n

4.17. (a) Because woman i is equally likely to be paired with any of the remaining

2n − 1 people, P(W

) =

2n−1

(b) Because, conditional on W

, woman i is equally likely to be paired with

any of 2n − 3 people, P(W

) =

2n−3

approximately be Poisson with mean



i=1

P(W

) =

2n−1

L 1/2. There-

fore, the probability that there is no such pairing is approximately e

−1/2

(d) It reduces to the match problem.

4.18. (a)





(9/19)

(10/19)

(9/19) =





(9/19)

(10/19)

(b) If W is her ﬁnal winnings and X is the number of bets she makes, then,

since she would have won 4 bets and lost X − 4 bets, it follows that

W = 20 − 5(X − 4) = 40 − 5X

Hence,

E[W] = 40 − 5E[X] = 40 − 5[4/(9/19)] =−20/9

4.19. The probability that a round does not result in an “odd person” is equal to 1/4,

the probability that all three coins land on the same side.

(a) (1/4)

(3/4) = 3/64

(b) (1/4)

= 1/256

Solutions to Self-Test Problems and Exercises 483

4.20. Let q = 1 − p. Then

E[1/X] =



i=1

i−1



i=1



i=1

i−1



i=1

i−1

1 − x

=−

log(p)

4.21. Since

X−b

a−b

will equal 1 with probability p or 0 with probability 1 − p, it follows

that it is a Bernoulli random variable with parameter p. Because the variance

of such a Bernoulli random variable is p(1 − p), we have

p(1 − p) = Var



X − b

a − b



(a − b)

Var(X − b) =

(a − b)

Var(X)

Hence,

Var(X) = (a − b)

p(1 − p)

4.22. Let X denote the number of games that you play and Y the number of games

that you lose.

(a) After your fourth game, you will continue to play until you lose. There-

fore, X − 4 is a geometric random variable with parameter 1 − p,so

E[X] = E[4 + (X − 4)] = 4 + E[X − 4] = 4 +

1 − p

(b) If we let Z denote the number of losses you have in the ﬁrst 4 games, then

Z is a binomial random variable with parameters 4 and 1 − p. Because

Y = Z + 1, we have

E[Y] = E[Z + 1] = E[Z] + 1 = 4(1 − p) + 1

4.23. A total of n white balls will be withdrawn before a total of m black balls if

and only if there are at least n white balls in the ﬁrst n + m − 1 withdrawals.

(Compare with the problem of the points, Example 4j of Chapter 3.) With X

equal to the number of white balls among the ﬁrst n + m − 1 balls withdrawn,

X is a hypergeometric random variable, and it follows that

484 Solutions to Self-Test Problems and Exercises

P{X Ú n}=

n+m−1



i=n

P{X = i}=

n+m−1



i=n





n + m − 1 − i





N + M

n + m − 1



4.24. Because each ball independently goes into urn i with the same probability p

,it

follows that X

is a binomial random variable with parameters n = 10, p = p

First note that X

+ X

is the number of balls that go into either urn i or

urn j. Then, because each of the 10 balls independently goes into one of these

urns with probability p

+ p

, it follows that X

+ X

is a binomial random

variable with parameters 10 and p

+ p

By the same logic, X

+ X

is a binomial random variable with param-

eters 10 and p

+ p

. Therefore,

P{X

+ X

= 7}=





( p

+ p

)

( p

+ p

)

4.25. Let X

equal 1 if person i has a match, and let it equal 0 otherwise. Then

X =



i=1

is the number of matches. Taking expectations gives

E[X] = E[



i=1

] =



i=1

E[X

] =



i=1

P{X

= 1}=



i=1

1/n = 1

where the ﬁnal equality follows because person i is equally likely to end up

with any of the n hats.

To compute Var(X), we use Equation (9.1), which states that

E[X

] =



i=1

E[X

] +



i=1



jZi

E[X

]

Now, for i Z j,

E[X

] = P{X

= 1, X

= 1}=P{X

= 1}P{X

= 1|X

= 1}=

n − 1

Hence,

E[X

] = 1 +



i=1



jZi

n(n − 1)

= 1 + n(n − 1)

n(n − 1)

= 2

which yields

Var(X) = 2 − 1

= 1

Solutions to Self-Test Problems and Exercises 485

4.26. With q = 1 − p, we have, on the one hand,

P(E) =



i=1

P{X = 2i}



i=1

2i−1

= pq



i=1

)

i−1

= pq

1 − q

(1 − q)(1 + q)

1 + q

On the other hand,

P(E) = P(E|X = 1)p + P(E|X > 1)q = qP(E|X > 1)

However, given that the ﬁrst trial is not a success, the number of trials needed

for a success is 1 plus the geometrically distributed number of additional trials

required. Therefore,

P(E|X > 1) = P(X + 1 is even) = P(E

) = 1 − P(E)

which yields P(E) = q/(1 + q).

CHAPTER 5

5.1. Let X be the number of minutes played.

(a) P{X > 15}=1 − P{X … 15}=1 − 5(.025) = .875

(b) P{20 < X < 35}=10(.05) + 5(.025) = .625

(d) P{X > 36}=4(.025)

= .1

5.2. (a) 1 =

dx = c/(n + 1) * c = n + 1

(b) P{X > x}=(n + 1)

dx = x

n+1

= 1 − x

n+1

5.3. First, let us ﬁnd c by using

1 =

dx = 32c/5 * c = 5/32

(a) E [X] =

dx =

= 5/3

(b) E[X

] =

dx =

128

= 20/7 * Var(X) = 20/7 − (5/3)

= 5/63

5.4. Since

1 =

(ax + bx

)dx = a/2 + b/3

.6 =

(ax

+ bx

)dx = a/3 + b/4

486 Solutions to Self-Test Problems and Exercises

we obtain a = 3.6, b =−2.4. Hence,

(a) P{X < 1/2}=

1/2

(3.6x − 2.4x

)dx = (1.8x

− .8x

)

1/2

= .35

(b) E[X

] =

(3.6x

− 2.4x

)dx = .42 * Var(X) = .06

5.5. For i = 1, ..., n,

P{X = i}=P{Int(nU) = i − 1}

= P{i − 1 … nU < i}

= P

i − 1

… U <

= 1/n

5.6. If you bid x,70 … x … 140, then you will either win the bid and make a proﬁt

of x − 100 with probability (140 − x)/70 or lose the bid and make a proﬁt of

0 otherwise. Therefore, your expected proﬁt if you bid x is

(x − 100)(140 − x) =

(240x − x

− 14000)

Differentiating and setting the preceding equal to 0 gives

240 − 2x = 0

Therefore, you should bid 120 thousand dollars. Your expected proﬁt will be

40/7 thousand dollars.

5.7. (a) P{U >.1}=9/10

(b) P{U >.2|U >.1}=P{U >.2}/P{U >.1}=8/9

(d) P{U >.3}=7/10

The answer to part (d) could also have been obtained by multiplying the prob-

abilities in parts (a), (b), and (c).

5.8. Let X be the test score, and let Z = (X − 100

)/15. Note that Z is a standard

normal random variable.

(a) P{X > 125}=P{Z > 25/15} L .0478

(b)

P{90 < X < 110}=P{−10/15 < Z < 10/15}

= P{Z < 2/3}−P{Z < −2/3}

= P{Z < 2/3}−[1 − P {Z < 2/3}]

L .4950

5.9. Let X be the travel time. We want to ﬁnd x such that

P{X > x}=.05

which is equivalent to

X − 40

x − 40

= .05

That is, we need to ﬁnd x such that

Z >

x − 40

= .05

Solutions to Self-Test Problems and Exercises 487

where Z is a standard normal random variable. But

P{Z > 1.645}=.05

Thus,

x − 40

= 1.645 or x = 51.515

Therefore, you should leave no later than 8.485 minutes after 12

P.M.

5.10. Let X be the tire life in units of one thousand, and let Z = (X − 34)/4. Note

that Z is a standard normal random variable.

(a) P{X > 40}=P{Z > 1.5} L .0668

(b) P{30 < X < 35}=P{−1 < Z <.25}=P{Z <.25}−P{Z > 1} L .44

(c)

P{X > 40|X > 30}=P{X > 40}/P{X > 30}

= P{Z > 1.5}/P{

Z > −1} L .079

5.11. Let X be next year’s rainfall and let Z = (X − 40.2)/8.4.

(a) P{X > 44}=P{Z > 3.8/8.4} L P{Z >.4524} L .3255

(b)





(.3255)

(.6745)

5.12. Let M

and W

denote, respectively, the numbers of men and women in the

samples that earn, in units of one thousand dollars, at least i per year. Also, let

Z be a standard normal random variable.

(a)

P{W

Ú 70}=P{W

Ú 69.5}

= P

− 200(.34)

√

200(.34)(.66)

69.5 − 200(.34)

√

200(.34)(.66)

L P{Z Ú .2239}

L .4114

(b)

P{M

… 120}=P{M

… 120.5}

= P

− (200)(.587)

√

(200)(.587)(.413)

…

120.5 − (200)(.587)

√

(200)(.587)(.413)

L P{Z … .4452}

L .6719

488 Solutions to Self-Test Problems and Exercises

(c)

P{M

Ú 150}=P{M

Ú 149.5}

= P

− (200)(.745)

√

(200)(.745)(.255)

149.5 − (200)(.745)

√

(200)(.745)(.255)

L P{Z Ú .0811}

L .4677

P{W

Ú 100}=P{W

Ú 99.5}

= P

− (200)(.534)

√

(200)(.534)(.466)

99.5 − (200)(.534)

√

(200)(.534)(.466)

L P{Z Ú −1.0348}

= L.8496

Hence,

P{M

Ú 150}P{W

Ú 100} L .3974

5.13. The lack of memory property of the exponential gives the result e

−4/5

5.14. (a) e

−2

= e

−4

(b) F(3) − F(1) = e

−1

− e

−9

−t

= 2t

(d) Let Z be a standard normal random variable. Use the identity E[X] =

P{X > x}dx to obtain

E[X] =

−x

= 2

−1/2

−y

= 2

−1/2

√

2πP{Z > 0}

√

π/2

(e) Use the result of Theoretical Exercise 5 to obtain

E[X

] =

2xe

−x

dx =−e

−x

= 1

Hence, Var(X) = 1 − π/4.

5.15. (a) P{X > 6}=exp{−

λ(t)dt}=e

−3.45

(b)

P{X < 8|X > 6}=1 − P{X > 8|X > 6}

= 1 − P{X > 8}/P{X > 6}

= 1 − e

−5.65

−3.45

L .8892

5.16. For x Ú 0,

1/X

(x) = P{1/X … x}

= P{X … 0}+P{X Ú 1/x}

= 1/2 + 1 − F

(1/x)

Solutions to Self-Test Problems and Exercises 489

Differentiation yields

1/X

(x) = x

−2

(1/x)

π(1 + (1/x)

)

= f

(x)

The proof when x < 0 is similar.

5.17. If X denotes the number of the ﬁrst n bets that you win, then the amount that

you will be winning after n bets is

35X − (n − X) = 36X − n

Thus, we want to determine

p = P{36X − n > 0}=P{X > n/36}

when X is a binomial random variable with parameters n and p = 1/38.

(a) When n = 34,

p = P{X Ú 1}

= P{X >.5} (the continuity correction)

= P

X − 34/38

34(1/38)(37/38)

.5 − 34/38

34(1/38)(37/38)

= P

X − 34/38

34(1/38)(37/38)

> −.4229

L (.4229)

L .6638

(Because you will be ahead after 34 bets if you win at least 1 bet, the exact

probability in this case is 1 − (37/38)

= .5961.)

(b) When n = 1000,

p = P{X > 27.5}

= P

X − 1000/38

1000(1/38)(37/38)

27.5 − 1000/38

1000(1/38)(37/38)

L 1 − (.2339)

L .4075

The exact probability—namely, the probability that a binomial n = 1000,

p = 1/38 random variable is greater than 27—is .3961.

490 Solutions to Self-Test Problems and Exercises

p = P{X > 2777.5}

= P

X − 100000/38

100000(1/38)(37/38)

2777.5 − 100000/38

100000(1/38)(37/38)

L 1 − (2.883)

L .0020

The exact probability in this case is .0021.

5.18. If X denotes the lifetime of the battery, then the desired probability,

P{X > s + t|X > t}, can be determined as follows:

P{X > s + t|X > t}=

P{X > s + t, X > t}

P{X > t}

P{X > s + t}

P{X > t}

P{X>s+t|battery is type 1}p

+P{X>s+t|battery is type 2}p

P{X>t|battery is type 1}p

+P{X>t|battery is type 2}p

−λ

(s+t)

+ e

−λ

(s+t)

−λ

+ e

−λ

Another approach is to directly condition on the type of battery and then

use the lack-of-memory property of exponential random variables. That is,

we could do the following:

P{X > s + t|X > t}=P{X > s + t|X > t,type1}P{type 1|X > t}

+ P{X > s + t|X > t,type2}P{type 2|X > t}

= e

−λ

P{type 1|X > t}+e

−λ

P{type 2|X > t}

Now for i = 1, 2, use

P{type i|X > t}=

P{type i, X > t}

P{X > t}

P{X > t|type i}p

P{X > t|type 1}p

+ P{X > t|type 2}p

−λ

+ e

−λ

5.19. Let X

be an exponential random variable with mean i, i = 1, 2.

(a) The value c should be such that P{X

> c}=.05. Therefore,

−c

= .05 = 1/20

or c = log(20) = 2.996.

(b)

P{X

> c}=e

−c/2

√

= .2236

Solutions to Self-Test Problems and Exercises 491

5.20. (a)

E[(Z − c)

] =

√

2π

−q

(x − c)

−x

√

2π

(x − c)e

−x

√

2π

−x

dx −

√

2π

−x

=−

√

2π

−x

− c(1 − (c))

√

2π

−c

− c(1 − (c))

(b) Using the fact that X has the same distribution as μ + σ Z, where Z is a

standard normal random variable, yields

E[(X − c)

] = E[(μ + σ Z − c)

]

= E

⎡

⎣





Z −

c − μ





⎤

⎦

= E



σ(Z −

c − μ

)



= σ E





Z −

c − μ





= σ



√

2π

−a

− a(1 − (a))



where a =

c−μ

CHAPTER 6

6.1. (a) 3C + 6C = 1 * C = 1/9

(b) Let p(i, j) = P{X = i, Y = j}. Then

p(1, 1) = 4/9, p(1, 0) = 2/9, P(0, 1) = 1/9, p(0, 0) = 2/9

(c)

(12)!

(1/9)

(2/9)

(d)

(12)!

(4!)

(1/3)

(e)



i=8





(2/3)

(1/3)

12−i

6.2. (a) With p

= P{XYZ = j}, we have

= p

= 1/4

Hence,

E[XYZ] = (6 + 2 + 4 + 12)/4 = 6

492 Solutions to Self-Test Problems and Exercises

(b) With q

= P{XY + XZ + YZ = j}, we have

= q

= 1/4

Hence,

E[XY + XZ + YZ] = (11 + 5 + 8 + 16)/4 = 10

6.3. In this solution, we will make use of the identity

−x

dx = n!

which follows because e

−x

/n!, x > 0, is the density function of a gamma

random variable with parameters n + 1andλ and must thus integrate to 1.

(a)

1 = C

−y

(y − x) dx dy

= C

−y

dy = 4C

Hence, C = 1/4.

(b) Since the joint density is nonzero only when y > x and y > −x, we have,

for x > 0,

(x) =

(y − x)e

−y

−(x+u)

−x

For x < 0,

(x) =

−x

(y − x)e

−y

[−ye

−y

− e

−y

+ xe

−y

]

−x

= (−2xe

+ e

)/4

(y) =

−y

(y − x) dx =

−y

(d)

E[X] =



−x

dx +

−q

(−2x

+ xe

) dx





1 −

(2y

−y

+ ye

−y

) dy



[1 − 4 − 1] =−1

(e) E [Y] =

−y

dy = 3

Solutions to Self-Test Problems and Exercises 493

6.4. The multinomial random variables X

, i = 1, ..., r, represent the numbers of

each of the types of outcomes 1, ..., r that occur in n independent trials when

each trial results in one of the outcomes 1, ..., r with respective probabili-

ties p

, ..., p

. Now, say that a trial results in a category 1 outcome if that

trial resulted in any of the outcome types 1, ..., r

; say that a trial results

in a category 2 outcome if that trial resulted in any of the outcome types

+ 1, ..., r

+ r

; and so on. With these deﬁnitions, Y

, ..., Y

represent

the numbers of category 1 outcomes, category 2 outcomes, up to category k

outcomes when n independent trials that each result in one of the categories

1, ..., k with respective probabilities



i−1

j=r

i−1

, i = 1, ..., k, are performed.

But by deﬁnition, such a vector has a multinomial distribution.

6.5. (a) Letting p

= P{XYZ = j}, we have

= 1/8, p

= 3/8, p

= 1/8

(b) Letting p

= P{XY + XZ + YZ = j}, we have

= 1/8, p

= 3/8, p

= 1/8

= P{X

+ YZ = j}, we have

= 1/8, p

= 1/4, p

= 1/8

6.6. (a)

1 =

(x/5 + cy) dy dx

(4x/5 + 12c) dx

= 12c + 2/5

Hence, c = 1/20.

(b) No, the density does not factor.

(c)

P{X + Y > 3}=

3−x

(x/5 + y/20) dy dx

[(2 + x)x/5 + 25/40 − (3 − x)

/40] dx

= 1/5 + 1/15 + 5/8 − 19/120 = 11/15

6.7. (a) Yes, the joint density function factors.

(b) f

(x) = x

ydy = 2x,0< x < 1

(y) = y

xdx = y/2, 0 < y < 2

(d)

P{X < x, Y < y}=P{X < x}P{Y < y}

= min(1, x

) min(1, y

/4), x > 0, y > 0

(e) E [Y] =

/2 dy = 4/3

494 Solutions to Self-Test Problems and Exercises

(f)

P{X + Y < 1}=

1−x

ydydx

x(1 − x)

dx = 1/24

6.8. Let T

denote the time at which a shock type i,ofi = 1, 2, 3, occurs. For

s > 0, t > 0,

P{X

> s, X

> t}=P{T

> s, T

> t, T

> max(s, t)}

= P{T

> s}P{T

> t}P{T

> max(s, t)}

= exp{−λ

s}exp{−λ

t}exp{−λ

max(s, t)}

= exp{−(λ

s + λ

t + λ

max(s, t))}

6.9. (a) No, advertisements on pages having many ads are less likely to be chosen

than are ones on pages with few ads.

(b)

n(i)

(c)



i=1

n(i)

= n/n, where n =



i=1

n(i)/m

(d) (1 −

n/n)

k−1

n(i)

= (1 −

n/n)

k−1

/(nm)

(e)



k=1

(1 −

n/n)

k−1

(f) The number of iterations is geometric with mean n

√

6.10. (a) P{X = i}=1/m, i = 1, ..., m.

(b) Step 2. Generate a uniform (0, 1) random variable U.IfU < n(X)/n,

go to step 3. Otherwise return to step 1.

Step 3. Generate a uniform (0, 1) random variable U, and select the

element on page X in position [n(X)U] + 1.

6.11. Yes, they are independent. This can be easily seen by considering the equiva-

lent question of whether X

is independent of N. But this is indeed so, since

knowing when the ﬁrst random variable greater than c occurs does not affect

the probability distribution of its value, which is the uniform distribution

on (c, 1).

6.12. Let p

denote the probability of obtaining i points on a single throw of the dart.

Then

= π/36

= 4π/36 − p

= π/12

= 9π/36 − p

− p

= 5π/36

= 1 − p

− p

= 1 − π/4

(a) π/12

(b) π/9

Solutions to Self-Test Problems and Exercises 495

(d) π(30/36 + 20/12 + 50/36) = 35π/9

(e) (π/4)

(f) 2(π/36)(1 − π/4) + 2(π/12)(5π/36)

6.13. Let Z be a standard normal random variable.

(a)

⎧

⎨

⎩



i=1

> 0

⎫

⎬

⎭

= P

⎧

⎪

⎨

⎪

⎩



i=1

− 6

√

−6

√

⎫

⎪

⎬

⎪

⎭

L P{Z > −1.2247} L .8897

(b)

⎧

⎨

⎩



i=1

> 0



i=1

=−5

⎫

⎬

⎭

= P{X

+ X

> 5}

= P

+ X

− 3

√

> 2/

√

L P{Z >.5774} L .2818

(c)

⎧

⎨

⎩



i=1

> 0|X

= 5

⎫

⎬

⎭

= P{X

+ X

> −5}

= P

+ X

− 4.5

√

> −9.5/

√

L P{Z > −2.239} L .9874

6.14. In the following, C does not depend on n.

P{N = n|X = x}=f

X|N

(x|n)P{N = n}/f

(x)

= C

(n − 1)!

(λx)

n−1

(1 − p)

n−1

= C(λ(1 − p)x)

n−1

/(n − 1)!

which shows that, conditional on X = x, N − 1 is a Poisson random variable

with mean λ(1 − p)x. That is,

P{N = n|X = x}=P{N − 1 = n − 1|X = x}

= e

−λ(1−p)x

(λ(1 − p)x)

n−1

/(n − 1)!, n Ú 1.

6.15. (a) The Jacobian of the transformation is

J =

= 1

As the equations u = x, v = x + y imply that x = u, y = v − u, we obtain

U,V

(u, v) = f

X,Y

(u, v − u) = 1, 0 < u < 1, 0 < v − u < 1

496 Solutions to Self-Test Problems and Exercises

or, equivalently,

U,V

(u, v) = 1, max(v − 1, 0)<u < min(v,1)

(b) For 0 < v < 1,

(v) =

du = v

For 1 … v … 2,

(v) =

v−1

du = 2 − v

6.16. Let U be a uniform random variable on (7, 11). If you bid x,7 … x … 10, you

will be the high bidder with probability

(P{U < x})



U − 7

x − 7





x − 7



Hence, your expected gain—call it E[G(x)]—if you bid x is

E[G(x)] =

(x − 7)

(10 − x)

Calculus shows this is maximized when x = 37/4.

6.17. Let i

, i

, ..., i

, be a permutation of 1, 2, ..., n. Then

P{X

= i

, X

= i

, ..., X

= i

}=P{X

= i

}P{X

= i

}···P{X

= i

}

= p

···p

= p

···p

Therefore, the desired probability is n! p

···p

, which reduces to

when

all p

= 1/n.

6.18. (a) Because



i=1



i=1

, it follows that N = 2M.

(b) Consider the n − k coordinates whose Y-values are equal to 0, and call

them the red coordinates. Because the k coordinates whose X-values are

equal to 1 are equally likely to be any of the





sets of k coordinates,

it follows that the number of red coordinates among these k coordinates

has the same distribution as the number of red balls chosen when one

randomly chooses k of a set of n balls of which n − k are red. Therefore,

M is a hypergeometric random variable.

2k(n−k)

(d) Using the formula for the variance of a hypergeometric given in

Example 8j of Chapter 4, we obtain

Var(N) = 4Var(M) = 4

n − k

n − 1

k(1 − k/n)(k/n)

6.19. (a) First note that S

− S



i=k+1

is a normal random variable with mean

0 and variance n − k that is independent of S

. Consequently, given that

= y, S

is a normal random variable with mean y and variance n − k.

Solutions to Self-Test Problems and Exercises 497

(b) Because the conditional density function of S

given that S

= x is a

density function whose argument is y, anything that does not depend on

y can be regarded as a constant. (For instance, x is regarded as a ﬁxed

constant.) In the following, the quantities C

, i = 1, 2, 3, 4 are all constants

that do not depend on y:

(y|x) =

(y, x)

(x)

= C

(x|y)f

(y)



where C

(x)



= C

√

2π

√

n − k

−(x−y)

/2(n−k)

√

2π

√

−y

/2k

= C

exp

−

(x − y)

2(n − k)

−

= C

exp

2xy

2(n − k)

−

2(n − k)

−

= C

exp

−

2k(n − k)



− 2



= C

exp

⎧

⎨

⎩

−

2k(n − k)





y −



−







⎫

⎬

⎭

= C

exp

−

2k(n − k)



y −



But we recognize the preceding as the density function of a normal ran-

dom variable with mean

x and variance

k(n − k)

6.20. (a)

P{X

> X

= max(X

, ..., X

)}

P{X

> X

, X

= max(X

, ..., X

)}

P{X

= max(X

, ..., X

)}

P{X

= max(X

, ..., X

), X

= max(X

, ..., X

)}

1/5

= 5

Thus, the probability that X

is the largest value is independent of which

is the largest of the other ﬁve values. (Of course, this would not be true if

the X

had different distributions.)

(b) One way to solve this problem is to condition on whether X

> X

. Now,

P{X

> X

= max(X

, ..., X

), X

> X

}=1

498 Solutions to Self-Test Problems and Exercises

Also, by symmetry,

P{X

> X

= max(X

, ..., X

), X

< X

From part (a),

P{X

> X

= max(X

, ..., X

)}=

Thus, conditioning on whether X

> X

yields the result

P{X

> X

= max(X

, ..., X

)}=

CHAPTER 7

7.1. (a) d =



i=1

1/n(i)

(b) P{X = i}=P{[mU] = i − 1}=P{i − 1 … mU < i}=1/m, i = 1, ..., m



n(X)





i=1

n(i)

P{X = i}=



i=1

n(i)

= d

7.2. Let I

equal 1 if the jth ball withdrawn is white and the (j + 1)st is black, and

let I

equal 0 otherwise. If X is the number of instances in which a white ball is

immediately followed by a black one, then we may express X as

X =

n+m−1



j=1

Thus,

E[X] =

n+m−1



j=1

E[I

]

n+m−1



j=1

P{j

selection is white, (j + 1)

is black}

n+m−1



j=1

P{j

selection is white}P{j + 1)

is black|j

is white}

n+m−1



j=1

n + m

n + m − 1

n + m

The preceding used the fact that each of the n + m balls is equally likely to

be the j th one selected and, given that that selection is a white ball, each of the

other n + m − 1 balls is equally likely to be the next ball chosen.

Solutions to Self-Test Problems and Exercises 499

7.3. Arbitrarily number the couples, and then let I

equal 1 if married couple num-

ber j, j = 1, ..., 10, is seated at the same table. Then, if X represents the

number of married couples that are seated at the same table, we have

X =



j=1

E[X] =



j=1

E[I

]

(a) To compute E[I

] in this case, consider wife number j. Since each of the





groups of size 3 not including her is equally likely to be the remain-

ing members of her table, it follows that the probability that her husband

is at her table is











Hence, E[I

] = 3/19 and so

E[X] = 30/19

(b) In this case, since the 2 men at the table of wife j are equally likely to be

any of the 10 men, it follows that the probability that one of them is her

husband is 2/10, so

E[I

] = 2/10 and E[X] = 2

7.4. From Example 2i, we know that the expected number of times that the die

need be rolled until all sides have appeared at least once is 6(1 + 1/2 + 1/3 +

1/4 + 1/5 + 1/6) = 14.7. Now, if we let X

denote the total number of times

that side i appears, then, since



i=1

is equal to the total number of rolls, we

have

14.7 = E

⎡

⎣



i=1

⎤

⎦



i=1

E[X

]

But, by symmetry, E[X

] will be the same for all i, and thus it follows from the

preceding that E[X

] = 14.7/6 = 2.45.

7.5. Let I

equal 1 if we win 1 when the jth red card to show is turned over, and let

equal 0 otherwise. (For instance, I

will equal 1 if the ﬁrst card turned over

is red.) Hence, if X is our total winnings, then

E[X] = E

⎡

⎢

⎣



j=1

⎤

⎥

⎦



j=1

E[I

]

500 Solutions to Self-Test Problems and Exercises

Now, I

will equal 1 if j red cards appear before j black cards. By symmetry, the

probability of this event is equal to 1/2; therefore, E[I

] = 1/2andE[X] = n/2.

7.6. To see that N … n − 1 + I, note that if all events occur, then both sides of

the preceding inequality are equal to n, whereas if they do not all occur, then

the inequality reduces to N … n − 1, which is clearly true in this case. Taking

expectations yields

E[N] … n − 1 + E[I]

However, if we let I

equal 1 if A

occurs and 0 otherwise, then

E[N] = E

⎡

⎣



i=1

⎤

⎦



i=1

E[I

] =



i=1

P(A

)

Since E[I] = P(A

···A

), the result follows.

7.7. Imagine that the values 1, 2, ..., n are lined up in their numerical order and that

the k values selected are considered special. From Example 3e, the position

of the ﬁrst special value, equal to the smallest value chosen, has mean 1 +

n − k

k + 1

n + 1

k + 1

For a more formal argument, note that X Ú j if none of the j − 1 smallest

values are chosen. Hence,

P{X Ú j}=



n − j + 1









n − k

j − 1





j − 1



which shows that X has the same distribution as the random variable of Exam-

ple 3e (with the notational change that the total number of balls is now n and

the number of special balls is k).

7.8. Let X denote the number of families that depart after the Sanchez family

leaves. Arbitrarily number all the N − 1 non-Sanchez families, and let I

1 … r … N − 1, equal 1 if family r departs after the Sanchez family does. Then

X =

N−1



r=1

Taking expectations gives

E[X] =

N−1



r=1

P{family r departs after the Sanchez family}

Now consider any non-Sanchez family that checked in k pieces of luggage.

Because each of the k + j pieces of luggage checked in either by this family or

by the Sanchez family is equally likely to be the last of these k + j to appear,

the probability that this family departs after the Sanchez family is

k+j

. Because

the number of non-Sanchez families who checked in k pieces of luggage is n

when k Z j,orn

− 1 when k = j, we obtain

E[X] =



k + j

−

Solutions to Self-Test Problems and Exercises 501

7.9. Let the neighborhood of any point on the rim be the arc starting at that point

and extending for a length 1. Consider a uniformly chosen point on the rim

of the circle—that is, the probability that this point lies on a speciﬁed arc of

length x is

2π

—and let X denote the number of points that lie in its neighbor-

hood. With I

deﬁned to equal 1 if item number j is in the neighborhood of the

random point and to equal 0 otherwise, we have

X =



j=1

Taking expectations gives

E[X] =



j=1

P{item j lies in the neighborhood of the random point}

But because item j will lie in its neighborhood if the random point is located

on the arc of length 1 going from item j in the counterclockwise direction, it

follows that

P{item j lies in the neighborhood of the random point}=

2π

Hence,

E[X] =

2π

> 3

Because E[X] > 3, at least one of the possible values of X must exceed 3,

proving the result.

7.10. If g(x) = x

1/2

,then



(x) =

−1/2

, g



(x) =−

−3/2

so the Taylor series expansion of

√

x about λ gives

√

X L

√

λ +

−1/2

(X − λ) −

−3/2

(X − λ)

Taking expectations yields

√

X] L

√

λ +

−1/2

E[X − λ] −

−3/2

E[(X − λ)

]

√

λ −

−3/2

√

λ −

−1/2

502 Solutions to Self-Test Problems and Exercises

Hence,

Var(

√

X) = E[X] − (E[

√

X])

L λ −



√

λ −

−1/2



= 1/4 −

64λ

L 1/4

7.11. Number the tables so that tables 1, 2, and 3 are the ones with four seats and

tables 4, 5, 6, and 7 are the ones with two seats. Also, number the women, and

let X

i,j

equal 1 if woman i is seated with her husband at table j. Note that

E[X

i,j

] =











, j = 1, 2, 3

and

E[X

i,j

] =





190

, j = 4, 5, 6, 7

Now, X denotes the number of married couples that are seated at the same

table, we have

E[X] = E

⎡

⎢

⎣



i=1



j=1

i,j

⎤

⎥

⎦



i=1



j=1

E[X

i,j

] +



i=1



j=4

E[X

i,j

]

7.12. Let X

equal 1 if individual i does not recruit anyone, and let X

equal 0 other-

wise. Then

E[X

] = P{i does not recruit any of i + 1, i + 2, ..., n}

i − 1

i + 1

···

n − 2

n − 1

i − 1

n − 1

Hence,

⎡

⎣



i=1

⎤

⎦



i=1

i − 1

n − 1

Solutions to Self-Test Problems and Exercises 503

From the preceding we also obtain

Var(X

) =

i − 1

n − 1



1 −

i − 1

n − 1



(i − 1)(n − i)

(n − 1)

Now, for i < j,

E[X

] =

i − 1

···

j − 2

j − 1

j − 2

j − 1

j + 1

···

n − 3

n − 1

(i − 1)(j − 2)

(n − 2)(n − 1)

Thus,

Cov(X

, X

) =

(i − 1)(j − 2)

(n − 2)(n − 1)

−

i − 1

n − 1

j − 1

n − 1

(i − 1)(j − n)

(n − 2)(n − 1)

Therefore,

Var

⎛

⎝



i=1

⎞

⎠



i=1

Var(X

) + 2

n−1



i=1



j=i+1

Cov(X

, X

)



i=1

(i − 1)(n − i)

(n − 1)

+ 2

n−1



i=1



j=i+1

(i − 1)(j − n)

(n − 2)(n − 1)

(n − 1)



i=1

(i − 1)(n − i)

−

(n − 2)(n − 1)

n−1



i=1

(i − 1)(n − i)(n − i − 1)

7.13. Let X

equal 1 if the ith triple consists of one of each type of player. Then

E[X

] =











Hence, for part (a), we obtain

⎡

⎣



i=1

⎤

⎦

= 6/7

It follows from the preceding that

Var(X

) = (2/7)(1 − 2/7) = 10/49

504 Solutions to Self-Test Problems and Exercises

Also, for i Z j,

E[X

] = P{X

= 1, X

= 1}

= P{X

= 1}P{X

= 1|X

= 1}





















= 6/70

Hence, for part (b), we obtain

Var

⎛

⎝



i=1

⎞

⎠



i=1

Var(X

) + 2



j>1

Cov(X

, X

)

= 30/49 + 2





−



312

490

7.14. Let X

, i = 1, ..., 13, equal 1 if the ith card is an ace and let X

be 0 otherwise.

Let Y

equal 1 if the jth card is a spade and let i, j = 1, ..., 13, be 0 otherwise.

Now,

Cov(X, Y) = Cov

⎛

⎜

⎝



i=1



j=1

⎞

⎟

⎠



i=1



j=1

Cov(X

, Y

)

However, X

is clearly independent of Y

because knowing the suit of a par-

ticular card gives no information about whether it is an ace and thus cannot

affect the probability that another speciﬁed card is an ace. More formally, let

i, s

, A

i, h

, A

i, d

, A

i, c

be the events, respectively, that card i is a spade, a heart, a

diamond, and a club. Then

P{Y

= 1}=

(P{Y

= 1|A

i,s

}+P{Y

= 1|A

i,h

}

+ P{Y

= 1|A

i,d

}+P{Y

= 1|A

i,c

})

But, by symmetry, we have

P{Y

= 1|A

i,s

}=P{Y

= 1|A

i,h

}=P{Y

= 1|A

i,d

}=P{Y

= 1|A

i,c

}

Therefore,

P{Y

= 1}=P{Y

= 1|A

i,s

}

As the preceding implies that

P{Y

= 1}=P{Y

= 1|A

i,s

}

Solutions to Self-Test Problems and Exercises 505

we see that Y

and X

are independent. Hence, Cov(X

, Y

) = 0, and thus

Cov(X, Y) = 0.

The random variables X and Y, although uncorrelated, are not indepen-

dent. This follows, for instance, from the fact that

P{Y = 13|X = 4}=0 Z P{Y = 13}

7.15. (a) Your expected gain without any information is 0.

(b) You should predict heads if p > 1/2 and tails otherwise.

E[Gain] =

E[Gain|V = p] dp

1/2

[1(1 − p) − 1(p)] dp +

1/2

[1(p) − 1(1 − p)] dp

= 1/2

7.16. Given that the name chosen appears in n(X) different positions on the list,

since each of these positions is equally likely to be the one chosen, it fol-

lows that

E[I|n(X)] = P{I = 1|n(X)}=1/n(X)

Hence,

E[I] = E[1/n(X)]

Thus, E[mI] = E[m/n(X)] = d.

7.17. Letting X

equal 1 if a collision occurs when the ith item is placed, and letting

it equal 0 otherwise, we can express the total number of collisions X as

X =



i=1

Therefore,

E[X] =



i=1

E[X

]

To determine E[X

], condition on the cell in which it is placed.

E[X

] =



E[X

| placed in cellj]p



P{i causes collision|placed in cell j]p



[1 − (1 − p

)

i−1

= 1 −



(1 − p

)

i−1

The next to last equality used the fact that, conditional on item i being placed

in cell j, item i will cause a collision if any of the preceding i − 1 items were

put in cell j. Thus,

506 Solutions to Self-Test Problems and Exercises

E[X] = m −



i=1



j=1

(1 − p

)

i−1

Interchanging the order of the summations gives

E[X] = m − n +



j=1

(1 − p

)

Looking at the result shows that we could have derived it more easily by taking

expectations of both sides of the identity

number of nonempty cells = m − X

The expected number of nonempty cells is then found by deﬁning an indicator

variable for each cell, equal to 1 if that cell is nonempty and to 0 otherwise,

and then taking the expectation of the sum of these indicator variables.

7.18. Let L denote the length of the initial run. Conditioning on the ﬁrst value gives

E[L] = E[L|ﬁrst value is one]

n + m

+ E[L|ﬁrst value is zero]

n + m

Now, if the ﬁrst value is one, then the length of the run will be the position of

the ﬁrst zero when considering the remaining n + m − 1 values, of which n − 1

are ones and m are zeroes. (For instance, if the initial value of the remaining

n + m − 1 is zero, then L = 1.) As a similar result is true given that the

ﬁrst value is a zero, we obtain from the preceding, upon using the result from

Example 3e, that

E[L] =

n + m

m + 1

n + m

n + 1

n + m

m + 1

n + 1

7.19. Let X be the number of ﬂips needed for both boxes to become empty, and let

Y denote the number of heads in the ﬁrst n + m ﬂips. Then

E[X] =

n+m



i=0

E[X|Y = i]P{Y = i}

n+m



i=0

E[X|Y = i]



n + m



(1 − p)

n+m−i

Now, if the number of heads in the ﬁrst n + m ﬂips is i, i … n, then the number

of additional ﬂips is the number of ﬂips needed to obtain an additional n − i

heads. Similarly, if the number of heads in the ﬁrst n + m ﬂips is i, i > n,

then, because there would have been a total of n + m − i < m tails, the

number of additional ﬂips is the number needed to obtain an additional i − n

heads. Since the number of ﬂips needed for j outcomes of a particular type is a

negative binomial random variable whose mean is j divided by the probability

of that outcome, we obtain

Solutions to Self-Test Problems and Exercises 507

E[X] =



i=0

n − i



n + m



(1 − p)

n+m−i

n+m



i=n+1

i − n

1 − p



n + m



(1 − p)

n+m−i

7.20. Taking expectations of both sides of the identity given in the hint yields

E[X

] = E



n−1

(x) dx



= n

E[x

n−1

(x)] dx

= n

n−1

E[I

(x)] dx

= n

n−1

F(x) dx

Taking the expectation inside the integral sign is justiﬁed because all the ran-

dom variables I

(x),0 < x < q, are nonnegative.

7.21. Consider a random permutation I

, ..., I

that is equally likely to be any of the

n! permutations. Then

E[a

j+1

] =



E[a

j+1

= k]P{I

= k}



E[a

j+1

= k]



P{I

j+1

= i|I

= k}

n(n − 1)



iZk

n(n − 1)



(−a

)

< 0

where the ﬁnal equality followed from the assumption that



i=1

= 0. Since

the preceding shows that

⎡

⎢

⎣



j=1

j+1

⎤

⎥

⎦

< 0

it follows that there must be some permutation i

, ..., i

for which



j=1

j+1

< 0

508 Solutions to Self-Test Problems and Exercises

7.22. (a) E[X] = λ

+ λ

, E [X] = λ

+ λ

(b)

Cov(X, Y) = Cov(X

+ X

, X

+ X

)

= Cov(X

, X

+ X

) + Cov(X

, X

+ X

)

= Cov(X

, X

)

= Var(X

)

= λ

gives

P{X = i, Y = j}=



P{X = i, Y = j|X

= k}P{X

= k}



P{X

= i − k, X

= j − k|X

= k}e

−λ

/k!



P{X

= i − k, X

= j − k}e

−λ

/k!



P{X

= i − k}P{X

= j − k}e

−λ

/k!

min(i,j)



k=0

−λ

i−k

(i − k)!

−λ

j−k

(j − k)!

−λ

7.23.

Corr

⎛

⎝



⎞

⎠

Cov(



)

Var(



)Var(



)



Cov(X

, Y

)

nσ



Cov(X

, Y

) +



jZi

Cov(X

, Y

)

nσ

nρσ

nσ

= ρ

where the next to last equality used the fact that Cov(X

, Y

) = ρσ

7.24. Let X

equal 1 if the ith card chosen is an ace, and let it equal 0 otherwise.

Because

X =



i=1

and E[X

] = P{X

= 1}=1/13, it follows that E[X] = 3/13. But, with A being

the event that the ace of spades is chosen, we have

Solutions to Self-Test Problems and Exercises 509

E[X] = E[X|A]P(A) + E[X|A

]P(A

)

= E[X|A]

+ E[X|A

]

= E[X|A]

⎡

⎣



i=1

⎤

⎦

= E[X|A]



i=1

E[X

]

= E[X|A]

Using that E[X] = 3/13 gives the result

E[X|A] =



−



= 1.1176

Similarly, letting L be the event that at least one ace is chosen, we have

E[X] = E[X|L]P(L) + E[X|L

]P(L

)

= E[X|L]P(L)

= E[X|L]



1 −

48 · 47 · 46

52 · 51 · 50



Thus,

E[X|L] =

3/13

1 −

48·47·46

52·51·50

L 1.0616

Another way to solve this problem is to number the four aces, with the ace of

spades having number 1, and then let Y

equal 1 if ace number i is chosen and

0 otherwise. Then

E[X|A] = E

⎡

⎣



i=1

= 1

⎤

⎦

= 1 +



i=2

E[Y

= 1]

= 1 + 3 ·

= 19/17

where we used that the fact given that the ace of spades is chosen the other

two cards are equally likely to be any pair of the remaining 51 cards; so the

conditional probability that any speciﬁed card (not equal to the ace of spades)

is chosen is 2/51. Also,

E[X|L] = E

⎡

⎣



i=1

⎤

⎦



i=1

E[Y

|L] = 4P{Y

= 1|L}

Because

P{Y

= 1|L}=P(A|L) =

P(AL)

P(L)

P(A)

P(L)

3/52

1 −

48·47·46

52·51·50

we obtain the same answer as before.

510 Solutions to Self-Test Problems and Exercises

7.25. (a) E[I|X = x] = P{Z < X|X = x}=P{Z < x|X = x}=P{Z < x}=(x)

(b) It follows from part (a) that E[I|X] = (X). Therefore,

E[I] = E[E[I|X]] = E[(X)]

The result now follows because E[I] = P{I = 1}=P{Z < X}.

X > Z}=P{X − Z > 0}

= P

X − Z − μ

−μ

= 1 − 



−μ



= 





7.26. Let N be the number of heads in the ﬁrst n + m − 1 ﬂips. Let M = max(X, Y)

be the number of ﬂips needed to amass at least n heads and at least m tails.

Conditioning on N gives

E[M] =



E[M|N = i]P{N = i}

n−1



i=0

E[M|N = i]P{N = i}+

n+m−1



i=n

E[M|N = i]P{N = i}

Now, suppose we are given that there are a total of i heads in the ﬁrst n+m − 1

trials. If i < n, then we have already obtained at least m tails, so the additional

number of ﬂips needed is equal to the number needed for an additional n − i

heads; similarly, if i Ú n, then we have already obtained at least n heads, so

the additional number of ﬂips needed is equal to the number needed for an

additional m − (n + m − 1 − i) tails. Consequently, we have

E[M] =

n−1



i=0



n + m − 1 +

n − i



P{N = i}

n+m−1



i=n



n + m − 1 +

i + 1 − n

1 − p



P{N = i}

= n + m − 1 +

n−1



i=0

n − i



n + m − 1



(1 − p)

n+m−1−i

n+m−1



i=n

i + 1 − n

1 − p



n + m − 1



(1 − p)

n+m−1−i

The expected number of ﬂips to obtain either n heads or m tails, E[min(X, Y)],

is now given by

E[min(X, Y)] = E[X + Y − M ] =

1 − p

− E[M]

Solutions to Self-Test Problems and Exercises 511

7.27. This is just the expected time to collect n − 1ofthen types of coupons in

Example 2i. By the results of that example the solution is

1 +

n − 1

n − 2

+ ... +

7.28. With q = 1 − p,

E[X] =



i=1

P{X Ú i}=



i=1

P{X Ú i}=



i=1

i−1

1 − q

7.29.

Cov(X, Y) = E[XY] − E[X]E[Y] = P(X = 1, Y = 1) − P(X = 1)P(Y = 1)

Hence,

Cov(X, Y) = 0 3 P(X = 1, Y = 1) = P(X = 1)P(Y = 1)

Because

Cov(X, Y) = Cov(1 − X,1 − Y) =−Cov(1 − X,

Y) =−Cov(X,1 − Y)

the preceding shows that all of the following are equivalent when X and Y are

Bernoulli:

1. Cov(X, Y) = 0

2. P(X = 1, Y = 1) = P(X = 1)P(Y = 1)

3. P(1 − X = 1, 1 − Y = 1) = P(1 − X = 1)P(1 − Y = 1)

4. P(1 − X = 1, Y = 1) = P (1 − X = 1)P(Y

= 1)

5. P(X = 1, 1 − Y = 1) = P(X = 1)P(1 − Y = 1)

7.30. Number the individuals, and let X

i,j

equal 1 if the jth individual who has hat

size i chooses a hat of that size, and let X

i,j

equal 0 otherwise. Then the number

of individuals who choose a hat of their size is

X =



i=1



j=1

i,j

Hence,

E[X] =



i=1



j=1

E[X

i,j

] =



i=1



j=1



i=1

CHAPTER 8

8.1. Let X denote the number of sales made next week, and note that X is integral.

From Markov’s inequality, we obtain the following:

(a) P{X > 18}=P{X Ú 19} …

E[X]

= 16/19

(b) P{X > 25}=P{X Ú 26} …

E[X]

= 16/26

512 Solutions to Self-Test Problems and Exercises

8.2. (a)

P{10 … X … 22}=P{|X − 16| … 6}

= P{|X − μ| … 6}

= 1 − P{|X − μ| > 6}

Ú 1 − 9/36 = 3/4

(b) P{X Ú 19}=P{X − 16 Ú 3} …

9 + 9

= 1/2

In part (a), we used Chebyshev’s inequality; in part (b), we used its

one-sided version. (See Proposition 5.1.)

8.3. First note that E[X − Y] = 0and

Var(X − Y) = Var(X) + Var(Y) − 2Cov(X, Y) = 28

Using Chebyshev’s inequality in part (a) and the one-sided version in parts (b)

and (c) gives the following results:

(a) P{|X − Y| > 15} … 28/225

(b) P{X − Y > 15} …

28 + 225

= 28/253

28 + 225

= 28/253

8.4. If X is the number produced at factory A and Y the number produced at fac-

tory B,then

E[Y − X] =−2, Var(Y − X) = 36 + 9 = 45

P{Y − X > 0}=P{Y − X Ú 1}=P{Y − X + 2 Ú 3} …

45 + 9

= 45/54

8.5. Note ﬁrst that

E[X

] =

dx = 2/3

Now use the strong law of large numbers to obtain

r = lim

n→q

= lim

n→q

lim

n→q

= 1/(2/3) = 3/2

8.6. Because E[X

] = 2/3and

E[X

] =

dx = 1/2

Solutions to Self-Test Problems and Exercises 513

we have Var(X

) = 1/2 − (2/3)

= 1/18. Thus, if there are n components on

hand, then

P{S

Ú 35}=P{S

Ú 34.5} (the continuity correction)

= P

− 2n/3

n/18

…

34.5 − 2n/3

n/18

L P

Z Ú

34.5 − 2n/3

n/18

where Z is a standard normal random variable. Since

P{Z > −1.284}=P{Z < 1.284} L .90

we see that n should be chosen so that

(34.5 − 2n/3) L −1.284

n/18

A numerical computation gives the result n = 55.

8.7. If X is the time required to service a machine, then

E[X] = .2 + .3 = .5

Also, since the variance of an exponential random variable is equal to the

square of its mean, we have

Var(X) = (.2)

+ (.3)

= .13

Therefore, with X

being the time required to service job i, i = 1, ..., 20, and Z

being a standard normal random variable, it follows that

P{X

+ ··· + X

< 8}=P

+ ··· + X

− 10

√

2.6

8 − 10

√

2.6

L P{Z < −1.24035}

L .1074

8.8. Note ﬁrst that if X is the gambler’s winnings on a single bet, then

E[X] =−.7 − .4 + 1 =−.1, E[X

] = .7 + .8 + 10 = 11.5

→Var(X) = 11.49

Therefore, with Z having a standard normal distribution,

P{X

+ ··· + X

100

… −.5}=P

+ ··· + X

100

+ 10

√

1149

…

−.5 + 10

√

1149

L P{Z … .2803}

L .6104

514 Solutions to Self-Test Problems and Exercises

8.9. Using the notation of Problem 7, we have

P{X

+ ··· + X

< t}=P

+ ··· + X

− 10

√

2.6

t − 10

√

2.6

L P

Z <

t − 10

√

2.6

Now, P{Z < 1.645} L .95, so t should be such that

t − 10

√

2.6

L 1.645

which yields t L 12.65.

8.10. If the claim were true, then, by the central limit theorem, the average nicotine

content (call it X) would approximately have a normal distribution with mean

2.2 and standard deviation .03. Thus, the probability that it would be as high

as 3.1 is

P{X > 3.1}=P

X − 2.2

√

.03

3.1 − 2.2

√

.03

L P{Z > 5.196}

L 0

where Z is a standard normal random variable.

8.11. (a) If we arbitrarily number the batteries and let X

denote the life of battery

i, i = 1, ..., 40, then the X

are independent and identically distributed

random variables. To compute the mean and variance of the life of, say,

battery 1, we condition on its type. Letting I equal 1 if battery 1 is type A

and letting it equal 0 if it is type B, we have

E[X

|I = 1] = 50 , E[X

|I = 0] = 30

yielding

E[X

] = 50P{I = 1}+30P{I = 0}=50(1/2) + 30(1/2) = 40

In addition, using the fact that E[W

] = (E[W])

+ Var(W), we have

E[X

|I = 1] = (50)

+ (15)

= 2725 , E[X

|I = 0] = (30)

+ 6

= 936

yielding

E[X

] = (2725)(1/2) + (936)(1/2) = 1830.5

Thus, X

, ..., X

are independent and identically distributed random

variables having mean 40 and variance 1830.5 − 1600 = 230.5. Hence,

with S =



i=1

, we have

E[S] = 40(40) = 1600 , Var(S) = 40(230.5) = 9220

Solutions to Self-Test Problems and Exercises 515

and the central limit theorem yields

P{S > 1700}=P

S − 1600

√

9220

1700 − 1600

√

9220

L P{Z > 1.041}

= 1 − (1.041) = .149

(b) For this part, let S

be the total life of all the type A batteries and let S

the total life of all the type B batteries. Then, by the central limit theorem,

has approximately a normal distribution with mean 20(50) = 1000 and

variance 20(225) = 4500, and S

has approximately a normal distribution

with mean 20(30) = 600 and variance 20(36) = 720. Because the sum of

independent normal random variables is also a normal random variable,

it follows that S

+ S

is approximately normal with mean 1600 and

variance 5220. Consequently, with S = S

+ S

P{S > 1700}=P

S − 1600

√

5220

1700 − 1600

√

5220

L P{Z > 1.384}

= 1 − (1.384) = .084

8.12. Let N denote the number of doctors who volunteer. Conditional on the event

N = i, the number of patients seen is distributed as the sum of i indepen-

dent Poisson random variables with common mean 30. Because the sum of

independent Poisson random variables is also a Poisson random variable, it

follows that the conditional distribution of X given that N = i is Poisson with

mean 30i. Therefore,

E[X|N] = 30N Var(X|N) = 30N

As a result,

E[X] = E[E[X|N]] = 30E[N] = 90

Also, by the conditional variance formula,

Var(X) = E[Var(X|N)] + Var(

E[X|N]) = 30E[N] + (30)

Var(N)

Because

Var(N) =

+ 3

+ 4

) − 9 = 2/3

we obtain Var(X) = 690.

To approximate P{X > 65}, we would not be justiﬁed in assuming that the

distribution of X is approximately that of a normal random variable with mean

90 and variance 690. What we do know, however, is that

P{X > 65}=



i=2

P{X > 65|N = i}P{N = i}=



i=2

(65)

516 Solutions to Self-Test Problems and Exercises

where P

(65) is the probability that a Poisson random variable with mean 30i

is greater than 65. That is,

(65) = 1 −



j=0

−30i

(30i)

/j!

Because a Poisson random variable with mean 30i has the same distribution

as does the sum of 30i independent Poisson random variables with mean 1,

it follows from the central limit theorem that its distribution is approximately

normal with mean and variance equal to 30i. Consequently, with X

being a

Poisson random variable with mean 30i and Z being a standard normal ran-

dom variable, we can approximate

(65) as follows:

(65) = P{X > 65}

= P{X Ú 65.5}

= P

X − 30i

√

30i

65.5 − 30i

√

30i

L P

Z Ú

65.5 − 30i

√

30i

Therefore,

(65) L P {Z Ú .7100} L .2389

(65) L P {Z Ú −2.583} L .9951

(65) L P {Z Ú −4.975} L 1

leading to the result

P{X > 65} L .7447

If we would have mistakenly assumed that X was approximately normal, we

would have obtained the approximate answer .8244. (The exact probability is

.7440.)

8.13. Take logarithms and then apply the strong law of large numbers to obtain

log

⎡

⎢

⎣

⎛

⎝



i=1

⎞

⎠

1/n

⎤

⎥

⎦



i=1

log(X

)→E[log(X

)]

Therefore,

⎛

⎝



i=1

⎞

⎠

1/n

→e

E[log(X

)]

CHAPTER 9

9.1. From axiom (iii), it follows that the number of events that occur between times

8 and 10 has the same distribution as the number of events that occur by time

2 and thus is a Poisson random variable with mean 6. Hence, we obtain the

following solutions for parts (a) and (b):

(a) P{N(10) − N(8) = 0}=e

−6

(b) E[N(10) − N(8)] = 6

Solutions to Self-Test Problems and Exercises 517

the process of events occurring is a Poisson process with rate λ. Hence,

the expected time of the ﬁfth event after 2

P.M. is 2 + E[S

] = 2 + 5/3.

That is, the expected time of this event is 3:40

P.M.

9.2. (a)

P{N(1/3) = 2|N(1) = 2}

P{N(1/3) = 2, N(1) = 2}

P{N(1) = 2}

P{N(1/3) = 2, N(1) − N(1/3) = 0}

P{N(1) = 2}

P{N(1/3) = 2}P{N(1) − N(1/3) = 0}

P{N(1) = 2}

(by axiom (ii))

P{N(1/3) = 2}P{N(2/3) = 0}

P{N(1) = 2}

(by axiom (iii))

−λ/3

(λ/3)

/2!e

−2λ/3

−λ

/2!

= 1/9

(b)

P{N(1/2) Ú 1|N(1) = 2}=1 − P{N(1/2) = 0|N(1) = 2}

= 1 −

P{N(1/2) = 0, N(1) = 2}

P{N(1) = 2}

= 1 −

P{N(1/2) = 0, N(1) − N(1/2) = 2}

P{N(1) = 2}

= 1 −

P{N(1/2) = 0}P{N(1) − N(1/2) = 2}

P{N(1) = 2}

= 1 −

P{N(1/2) = 0}P{N(1/2) = 2}

P{N(1) = 2}

= 1 −

−λ/2

(λ/2)

/2!

−λ

/2!

= 1 − 1/4 = 3/4

9.3. Fix a point on the road and let X

equal 0 if the nth vehicle to pass is a car

and let it equal 1 if it is a truck, n Ú 1. We now suppose that the sequence

, n Ú 1, is a Markov chain with transition probabilities

0,0

= 5/6, P

0,1

= 1/6, P

1,0

= 4/5, P

1,1

= 1/5

Then the long-run proportion of times is the solution of

= π

(5/6) + π

(4/5)

= π

(1/6) + π

(1/5)

+ π

= 1

Solving the preceding equations gives

= 24/29 π

= 5/29

Thus, 2400/29 L 83 percent of the vehicles on the road are cars.

518 Solutions to Self-Test Problems and Exercises

9.4. The successive weather classiﬁcations constitute a Markov chain. If the states

are 0 for rainy, 1 for sunny, and 2 for overcast, then the transition probability

matrix is as follows:

P =

01/21/2

1/31/31/3

The long-run proportions satisfy

= π

(1/3) + π

(1/3)

= π

(1/2) + π

(1/3) + π

(1/3)

= π

(1/2) + π

(1/3) + π

(1/3)

1 = π

+ π

The solution of the preceding system of equations is

= 1/4, π

= 3/8, π

= 3/8

Hence, three-eighths of the days are sunny and one-fourth are rainy.

9.5. (a) A direct computation yields

H(X)/H(Y) L 1.06

(b) Both random variables take on two of their values with the same prob-

abilities .35 and .05. The difference is that if they do not take on either

of those values, then X, but not Y, is equally likely to take on any of its

three remaining possible values. Hence, from Theoretical Exercise 13, we

would expect the result of part (a).

CHAPTER 10

10.1. (a)

1 = C

dx * C = 1/(e − 1)

(b)

F(x) = C

dy =

− 1

e − 1

,0… x … 1

Hence, if we let X = F

−1

(U),then

U =

− 1

e − 1

X = log(U(e − 1) + 1)

Thus, we can simulate the random variable X by generating a random

number U and then setting X = log(U(e − 1) + 1).

10.2. Use the acceptance–rejection method with g(x) = 1, 0 < x < 1. Calculus shows

that the maximum value of f(x)/g(x) occurs at a value of x,0< x < 1, such that

2x − 6x

+ 4x

= 0

Solutions to Self-Test Problems and Exercises 519

or, equivalently, when

− 6x + 2 = (4x − 2)(x − 1) = 0

The maximum thus occurs when x = 1/2, and it follows that

C = max f (x)/g(x) = 30(1/4 − 2/8 + 1/16) = 15/8

Hence, the algorithm is as follows:

Step 1. Generate a random number U

Step 2. Generate a random number U

Step 3. If U

… 16(U

− 2U

+ U

), set X = U

; else return to Step 1.

10.3. It is most efﬁcient to check the higher probability values ﬁrst, as in the follow-

ing algorithm:

Step 1. Generate a random number U.

Step 2. If U … .35, set X = 3 and stop.

Step 3. If U … .65, set X = 4 and stop.

Step 4. If U … .85, set X = 2 and stop.

Step 5. X = 1.

10.4. 2μ − X

10.5. (a) Generate 2n independent exponential random variables with mean 1, X

, Y

, i =

1, ..., n, and then use the estimator



i=1

/n.

(b) We can use XY as a control variate to obtain an estimator of the type



i=1

+ cX

)/n

Another possibility would be to use XY + X

/2 as the control variate

and so obtain an estimator of the type



i=1

+ c[X

+ X

/2 − 1/2])/n

The motivation behind the preceding formula is based on the fact that the

ﬁrst three terms of the MacLaurin series expansion of e

are 1 + xy +

)/2.

This page intentionally left blank

Index

Absolutely continuous random

variables, See Continuous

random variables

Algorithm, polar, 453

Analytical Theory of Probability

(Laplace), 399

Answers to selected problems,

456–457

Antithetic variables, variance

reduction, 450–451

Archimedes, 208

Ars Conjectandi (The Art of

Conjecturing), 142, 391

Associative laws, 25

Axiom, deﬁned, 27

Axioms of probability, 26–29

Banach match problem, 158–159

Basic principle of counting, 1–3

proof of, 2

Bayes, Thomas, 74

Bayes’s formula, 65–79, 101

Bell, E. T., 208

Bernoulli, Jacques, 142–143

Bernoulli, James, 134, 143, 391

Bernoulli, Nicholas, 391

Bernoulli random variables,

134–139, 403

Bernoulli trials, 112

Bernstein polynomials, 414

Bertrand, Joseph L. F., 197

Bertrand’s paradox, 197

Best-prize problem, 344–346

Beta distribution, 218–219

Binary symmetric channel, 433

Binomial coefﬁcients, 7, 15

Binomial distribution, normal

approximation to, 204–207

Binomial random variables,

134–139, 259–260

binomial distribution function,

computing, 142–143

moments of, 316–317

properties of, 139–141

simulating, 448

variance of, 325–331

Binomial theorem, 7

combinatorial proof of, 8–9

proof by mathematical

induction, 8

Bits, 426

Bivariate normal distribution,

268–269

Boole’s inequality, 300–301

Borel,

E., 403

Box–Muller approach, 445

Branching process, 383

Buffon’s needle problem,

243–246

Cantor distribution, 381

Cauchy distribution, 217–218

Center of gravity, 128

Central limit theorem, 198, 412

Channel capacity, 434, 436

Chapman–Kolmogorov equations,

421–423

Chebyshev’s inequality:

deﬁned, 389

one-sided, 403–407

and weak law of large numbers,

388–391

Chernoff bounds, 407–409

Chi-squared distribution, 216, 255

Chi-squared random variable,

simulating, 446–447

Coding theory:

binary symmetric channel, 433

and entropy, 428–434

noiseless coding theorem,

429–431, 433–434

521

522 Index

Combinatorial analysis, 1–21

combinations, 5–9

integer solutions of equations,

number of, 12–15

multinomial coefﬁcients, 9–12

permutations, 3–5

principle of counting, 1–3

Combinatorial identity, 7

Commutative laws, 25

Complement, 24, 49

Conditional covariance formula,

372, 381

Conditional distributions:

continuous case, 266–274

bivariate normal distribution,

268–269

discrete case, 263–266

Conditional expectation, 331–349,

371

computing expectations by

conditioning, 333–343

computing probabilities by

conditioning, 344–347

best-prize problem, 344–346

conditional variance, 347–349

deﬁnitions, 331–333

and prediction, 349–353

Conditional independence, 98

Conditional probability, 58–65

Bayes’s formula, 65–79

independent events, 79–93

Conditional probability density

function, 286

Conditional probability mass

function, 286

Conditional variance, 347–349

variance of a sum of a random

number of random

variables, 349

Conditional variance formula, 372

Conditioning:

computing expectations by,

333–343

computing probabilities by,

344–347

best-prize problem, 344–346

variance reduction by, 451–452

Continuity correction, 205

Continuous random variables,

186–231

beta distribution, 218–219

Cauchy distribution, 217–218

expectation of, 190–194

gamma distribution, 215–216

simulation of:

general techniques for,

440–447

inverse transformation

method, 441–442, 453

rejection method, 442–444,

453–454

Weibull distribution, 216–217

Control variates, variance reduction,

452–453

Convolution, 252

Correlation, 371

Correlation coefﬁcient, 322–331

Counting, basic principle of, 1–3

proof of, 2

Coupon-collecting problem,

318–319

singletons in, 321–322

Coupon collecting with unequal

probabilities, 314–315

Covariance, 322–323, 371

deﬁned, 322

Cumulative distribution function

(distribution function),

123–125

properties, 168–170

DeMoivre, Abraham, 198, 207–208,

393

DeMoivre–Laplace limit theorem,

204

DeMorgan’s laws, 26

Dependent random variables, 241

Deviations, 324

Discrete distributions:

simulation from, 447–449

binomial random variable,

448

Index 523

geometric distribution,

447–448

Poisson random variable, 449

Discrete probability distribution,

358

Discrete random variables, 123–125,

171

Distribution function (distribution

function), 123, 170

Distributions:

beta, 218-219

binomial, normal

approximation to, 204-207

bivariate normal, 268-269

Cantor, 381

Cauchy, 217-218

chi-squared, 216, 255

conditional, 263-274

continuous probability, 359

discreet, 447-449

discrete probability, 358

gamma, 215-216

Gaussian, 207

geometric, 447-448

Laplace, 211

marginal, 233

mulltinomial, 240

multivariate, 372

multivariate normal, 365-367

n-Erlang, 216

negative hypergeometric, 319

normal, 356-357

Weibull, 216-217

zeta (Zipf), 163–164

Distributive laws, 25

Double exponential random

variable, 211fn

Duration of play, problem of, 89–90

Edges, 91

Ehrenfest, Paul and Tatyana, 421

Entropy, 426–428, 435

and coding theory, 428–434

Equations, number of integer

solutions of, 12–15

Ergodic Markov chain, 423–424

Events, 23–26

independent, 79–93, 101

mutually exclusive, 24, 49

odds of, 101

pairwise independent, 147

Exchangeable random variables,

282–285

Expectation, See Continuous

random variables

conditional, 331–349, 371

and prediction, 349–353

correlations, 322–331

covariance, 322–323

general deﬁnition of, 369–370

moment generating functions,

354–365

binomial distribution with

parameters n and p, 355

continuous probability

distribution, 359

determination of the

distribution, 358

discrete probability

distribution, 358

exponential distribution with

parameter λ, 356

independent binomial random

variables, sums of, 360

independent normal random

variables, sums of, 360

independent Poisson random

variables, sums of, 360

joint, 363–365

normal distribution, 356–357

Poisson distribution with

mean λ, 355–356

of the sum of a random

number of random

variables, 361–363

moments of the number of

events that occur, 315–322,

319

binomial random variables,

moments of, 316–317

coupon-collecting problem,

318–319, 321–322

524 Index

Expectation, See Continuous

random variables

(Continued)

hypergeometric random

variables, moments of,

317–318

moments in the match

problem, 318

negative hypergeometric

random variables, 319–321

probabilistic method, obtaining

bounds from expectations

via, 311–312

properties of, 297–387

of sums of random variables,

298–315

Boole’s inequality, 300–301

coupon-collecting problems,

303

coupon collecting with

unequal probabilities,

314–315

expectation of a binomial

random variable, 301

expected number of matches,

303

expected number of runs,

304–305

hypergeometric random

variable, mean of, 302

maximum–minimums

identity, 313–314

negative binomial random

variable, mean of, 301–302

probability of a union of

events, 308–310

quick-sort algorithm,

analyzing, 306–308

random walk in a plane,

305–306

sample mean, 300

variance of sums, 322–331

Expected value (expectation),

125–128, 171

Exponential random variables,

208–214, 223

hazard rate functions, 212–214

Failure rate function, 212, 223

Fermat, Pierre de, 85–86, 89

Fermat’s combinatorial identity, 18

Finite population, sampling from,

326–331

First of moment of X, 132

Functional system, 1

Galton, Francis, 399

Gambler’s ruin problem, 87–88

Gamma distribution, 215–216

Gamma function, 215, 223

Gamma random variables,

254–255

Gauss, Karl Friedrich, 207–208

Gaussian curve, 207

Gaussian distribution, 207

Generalized basic principle of

counting, 2

Geometric distribution, 447–448

variance of, 340

Geometric random variable with

parameter p, 448

Geometric random variables,

155–157, 260–263

Geometrical probability, 197

Goodwill cost, deﬁned, 176

Half-life, probabilistic interpretation

of (example), 249–251

Hamiltonian path:

deﬁned, 311

maximum number of, in a

tournament, 311–312

Hazard rate functions, 212–214, 223

Huygens, Christiaan, 86, 89

Hypergeometric random variables,

160–163

moments of, 317–318

Identically distributed uniform

random variables, 252–254

Importance sampling, 455

Index 525

Independent Bernoulli random

variables, bounding error

probability when

approximating a sum of,

410–412

Independent binomial random

variables, sums of, 260, 360

Independent events, 79–93, 101

Independent increment assumption,

417

Independent normal random

variables, sums of, 360

Independent Poisson random

variables, sums of, 259–260,

360

Independent random variables,

240–251, 286

binomial random variables,

259–260

Buffon’s needle problem,

243–246

conditional distributions:

continuous case, 266–274

discrete case, 263–266

gamma random variables,

254–255

geometric random variables,

260–263

half-life, probabilistic

interpretation of (example),

249–251

identically distributed uniform

random variables, 252–254

normal random variables,

256–259

Poisson random variables,

259–260

random subsets, 246–249

sums of, 252–263

Independent uniform random

variables, sum of two,

252–253

Inequality:

Boole’s, 300–301

Chebyshev’s, 388–391

Jensen’s, 409

Markov’s, 388

Initial probabilities, 99

Integer solutions of equations,

number of, 12–15

Interarrival times, sequence of,

418

Intersection, 23–24, 49

Inverse transformation method,

441–442, 453

exponential random variable,

simulating, 441

gamma random variable,

simulating, 441–442

Jacobian determinant, 280

Jensen’s inequality, 409

Joint cumulative probability

distribution function,

232–240, 282–285

Joint density function of order

statistics, 270

Joint distribution functions,

232–240

joint cumulative probability

distribution function,

232–240, 282–285

joint probability mass function,

233–234

marginal distributions, 233

multinomial distribution, 240

Joint moment generating functions,

363–365

Joint probability density function,

235–239, 285

Joint probability mass function,

233–234

Jointly continuous random

variables, 235, 239, 285

Jointly distributed random

variables, 232–297

joint distribution functions,

232–240

joint probability density

function, 235–239, 285

marginal probability mass

functions, 234

526 Index

Kelley strategy, 378

Khinchine, A. Y., 391

Kolmogorov, A. N., 403

Laplace distribution, 211

Laplace, Pierre-Simon, 399

Laplace’s rule of succession, 98–99

Law of frequency of errors, 399

Laws of large numbers, 388

Legendre theorem, 229

Length of the longest run

(example), 148–154

L’H

opital’s rule, 392

Liapounoff, A., 393

Limit theorems, 388–417

central limit theorem, 391–399,

412

Chebyshev’s inequality, 388–391

strong law of large numbers,

400–403, 412

weak law of large numbers,

388–391

Lognormal random variable, 258

Marginal distributions, 233

Marginal probability mass functions,

234

Markov chain, 419–424, 434

Chapman–Kolmogorov

equations, 421–422

ergodic, 423–424

matrix, 420

random walk, 422

transition probabilities, 420

Markov’s inequality, 388

Matching problem (example),

41–42, 63

Matrix, 420

Maximum likelihood estimate, 160

Maximum–minimums identity,

313–314

Mean, 132, 171

Measureable events, 29

Measurement signalto-noise ratio,

414

Memoryless, use of term, 223

Men of Mathematics (Bell), 208

Method of maximum likelihood

estimation, 180

Midrange of a sequence, 293

Minimax theorem, 175

Moment generating functions,

354–365, 372

binomial distribution with

parameters n and p, 355

continuous probability

distribution, 359

determination of the

distribution, 358

discrete probability distribution,

358

exponential distribution with

parameter λ, 356

independent binomial random

variables, sums of, 360

independent normal random

variables, sums of, 360

independent Poisson random

variables, sums of, 360

joint, 363–365

normal distribution, 356–357

Poisson distribution with mean

λ, 355–356

of the sum of a random number

of random variables,

361–363

Multinomial coefﬁcients, 9–12

deﬁned, 11

Multinomial distribution, 240

Multinomial theorem, 11

Multiplication rule, 62–63, 101

Multivariate distributions, 372

Mutually exclusive events, 24, 49

n-Erlang distribution, 216

Natural Inheritance (Galton), 399

Negative binomial random

variables, 157–160

Negative hypergeometric

distribution, 319

Index 527

Negative hypergeometric random

variables, 319–321

Newton, Isaac, 208

Noiseless coding theorem, 429–431,

433–434

Normal curve, 207

Normal random variables,

256–259

joint distribution of the sample

mean and sample variance,

367–369

multivariate normal

distribution, 365–367

simulating, 443–444

polar method, 445–446

Notation/terminology, 6, 10

nth moment of X, 132

Null event, 24

Null set, 49

Odds, of an event, 101

One-sided Chebyshev’s inequality,

403–407

Order statistics, 270–274, 286

distribution of the range of a

random sample, 273–274

joint density function of, 270

Overlook probabilities, 74

Pairwise independent events, 147

Pareto, V., 164

Pascal, Blaise, 85–86

Pearson, Karl, 207–208

Permutations, 3–5

Personal view of probability, 48

Pierre-Simon, Marquis de Laplace,

399

Points, problem of, 86

Poisson distribution function,

computing, 154–155

Poisson paradigm, 148

Poisson process, 417–419

deﬁned, 417, 434

independent increment

assumption, 417

sequence of interarrival times,

418

stationary increment

assumption, 417

waiting time, 419

Poisson random variables, 143–145,

171, 259–260

simulating, 449

Poisson, Sim

eon Denis, 143

Polar algorithm, 453

Polar method, 445–446

Polya’s urn model, 284

Posterior probability, 99–100

Principle of counting, 1–3

Prior probabilities, 99

Probabilistic method, 93

obtaining bounds from

expectations via, 311–312

maximum number of

Hamiltonian paths in a

tournament, 311–312

Probabilistic Method, The

(Alon/Spencer/Erdos), 93fn

Probability:

axioms of, 26–29

as a continuous set function,

44–48

deﬁning, 26

of an event, 27

geometrical, 197

as a measure of belief, 48–49

multiplication rule, 62–63, 101

personal view of, 48

sample space and events, 22–26

simple propositions, 29–33

subjective view of, 48

Probability density function, 222

deﬁned, 186

Probability mass function, 123, 171,

233

Probability, personal view of, 48

Problem of duration of play, 89–90

Pseudorandom numbers, 439

Quick-sort algorithm, analyzing,

306–308

528 Index

Random-number generator, 439

seed, 439

Random numbers, 385, 439, 453

Random permutation, generating,

439–440

Random samples, distribution of the

range of, 273–274

Random variables, 117–186, 134–139

Bernoulli, 134–139

binomial, 134–139, 259–260

continuous, 186–231

cumulative distribution

function, 123–125

properties of, 168–170

deﬁned, 117, 170

dependent, 241

discrete, 123–125, 171

distribution of a function of,

219–221

exchangeable, 282–285

expectation of a function of,

128–132

expectation of a sum of a

random number of, 335

expectation of sums of, 298–315

expected value (expectation),

125–128

sums of, 164–168

exponential, 208

gamma, 254–255

geometric, 155–157, 260–263

hypergeometric, 160–163

Identically distributed uniform,

252–254

independent, 240–251

joint probability distribution of

functions of, 274–282

jointly continuous, 239

moment generating functions,

354–365

of the sum of a random

number of, 361–363

negative binomial, 157–160

normal, 198–204, 256–259

order statistics, 270–274, 286

Poisson, 143–145, 171, 259–260

uniform, 194–198

variance, 171

variance of a sum of a random

number of, 349

Weibull, 217

zeta (Zipf) distribution, 163–164

Random walk, 422

Rate of transmission of information,

436

Rayleigh density function, 229

Recherches sur la probabilil

edes

jugements en mati

ere

criminelle et en mati

ere

civile (Investigations into the

Probability of Verdicts in

Criminal and Civil Matters),

143

Rejection method, 442–444, 453–454

simulating a chi-squared

random variable, 446–447

simulating a normal random

variable, 443–444

polar method, 445–446

Riemann, G. F. B., 164

Sample mean, 300, 372

joint distribution of, 367–369

joint distribution of sample

variance and, 367–369

Sample median, 272

Sample spaces:

and events, 22–26, 49

having equally likely outcomes,

33–44

Sample variance, 324, 372

joint distribution of sample

mean and, 367–369

Seed, 439

Selected problems, answers to,

456–457

Self-text problems/exercises,

458–516

Sequence of interarrival times, 418

Sequential updating of information,

99–101

Shannon, Claude, 433

Index 529

Simulation, 438–456

of continuous random variables:

general techniques for,

440–447

inverse transformation

method, 441–442, 453

rejection method, 442–444,

453–454

deﬁned, 438

from discrete distributions,

447–449

pseudorandom numbers, 439

random-number generator, 439

random numbers, 439, 453

random permutation,

generating (example),

439–440

seed, 439

variance reduction techniques,

449–453

Simulation, deﬁned, 246

Singletons, in coupon-collecting

problem, 321–322

Size of the zeroth generation, 383

St. Petersburg paradox, 175

Standard deviation of X, 134, 171

Standard deviations, 414

Stationary increment assumption,

417

Stieltjes integrals, 369–370

Strong law of large numbers,

400–403, 412

Subjective view of probability, 48

Sums of random variables:

expectation of, 298–315

binomial random variable,

300–301

Boole’s inequality, 300–301

coupon-collecting problems,

303

coupon collecting with

unequal probabilities,

314–315

expectation of a binomial

random variable, 301

expected number of matches,

303

expected number of runs,

304–305

hypergeometric random

variable, mean of, 302

maximum–minimums

identity, 313–314

negative binomial random

variable, mean of, 301–302

probability of a union of

events, 308–310

quick-sort algorithm,

analyzing, 306–308

random walk in a plane,

305–306

sample mean, 300

Superset, 24

Surprise concept, 425–426

Theory of games, 175

Transition probabilities, Markov

chains, 420

Trials, 82

Uncertainty, 426–427

Uniform random variables, 194–198

Union, 23–24, 49

Updated probability, 99–100

Updating information sequentially,

99–101

Utility, 130–131

Variables, 163–164, See also

Random variables

antithetic, 450–451

Variance, 171

conditional, 347–349

covariance, 322–323

of geometric distribution, 340

sample, 324, 372

Variance reduction:

antithetic variables, use of,

450–451

by conditioning, 451–452

control variates, 452–453

techniques, 449–453

530 Index

Venn diagram, 24–25

Vertices, 91

Waiting time, 419

Weak law of large numbers, 388–391

Weibull distribution, 216–217

Weibull random variables, 217

Wheel of fortune game

(chuck-a-luck) (example),

136

Wilcoxon sum-of ranks test, 376

Zeroth generation, size of, 383

Zeta (Zipf) distribution, 163–164

Zipf, G. K., 164